Vector Algebra Class 12 Maths Important Questions Chapter 10

Vector Algebra Class 12 Important Questions with Solutions Previous Year Questions

Question 1.
Find the position vector of a point which divides the join of points with position vectors a⃗ −2b⃗  and 22a⃗ +b⃗  externally in the ratio 2:1. (Delhi 2016)
Answer:
Let given position vectors are OA−→−=a⃗ −2b⃗  and OB−→−=2a⃗ +b⃗ .

Let OA−→− be the position vector of a point C which divides the join of points, with position vectors
OA−→− and OB−→−, externally in the ratio 2:1.
∴ OA−→− = 2OB−→−−1OA−→−2−1=2(2a⃗ +b⃗ )−1(a⃗ −2b⃗ )1 [by external section formula]
= 4a⃗  + 2b⃗  – a⃗  + 2b⃗  = 3a⃗  + 4b⃗ 

Question 2.
If a⃗  = 4î – ĵ + k̂ and b⃗  = 2î – 2ĵ + k̂, then find a unit vector parallel to the vector a⃗ +b⃗ . (All India 2016)
Answer:
Given vectors are
a⃗  = 4î – ĵ + k̂, b⃗  = 2î – 2ĵ + k̂.
Now, a⃗ +b⃗  == (4î – ĵ + k̂) + (2î – 2ĵ + k̂)
= 6î – 3ĵ + 2k̂

and |a⃗ +b⃗ |=(6)2+(−3)2+(2)2−−−−−−−−−−−−−−−√
= 36+9+4−−−−−−−−√=49−−√ = 7units

∴ The unit vector parallelto the vector a⃗ +b⃗  is
a⃗ +b⃗ |a⃗ +b⃗ |=6i^−3j^+2k^7

Question 3.
The two vectors ĵ + k̂ and 3î – ĵ + 4k̂ represent the two sides AB−→− and AC−→− respectively of triangle ABC. Find the length of the median through A. (Delhi 2016; Foreign 2015)
Answer:
Given, AB−→− = ĵ + k̂ and AC−→− = 3î – ĵ + 4k̂

Alternate Method:
Given AB−→− = ĵ + k̂ and AC−→− = 3î – ĵ + 4k̂

Question 4.
Write the direction ratios of the vector 3a⃗  + 2b⃗ , where a⃗  = î + ĵ – 2k̂ and b⃗  = 2î – 4ĵ + 5k̂ (All India 2015C)
Answer:
Clearly, 3a⃗  + 2b⃗  = 3 (î + ĵ – 2k̂) + 2 (2î – 4ĵ + 5k̂)
= (3î + 3ĵ – 6k̂) + (4î – 8ĵ + 10k̂)
= 7î – 5ĵ + 4k̂
Hence, direction ratios of vectors 3a⃗  + 2b⃗  are 7, – 5 and 4.

Question 5.
Find the unit vector in the direction of the sum of the vectors 2î + 3ĵ – k̂ and 4î – 3ĵ + 2k̂. (Foreign 2015)
Answer:
Let a⃗  = 2î + 3ĵ – k̂ and b⃗  = 4î – 3ĵ + 2k̂
Now, sum of two vectors,
a⃗ +b⃗  = (2î + 3ĵ – k̂) + (4î – 3ĵ + 2k̂) = 6î + k̂

Question 6.
Find a vector in the direction of vector 2î – 3ĵ + 6k̂ which has magnitude 21 units. (Foreign 2014)
Answer:
To find a vector in the direction of given vector, first of all we find unit vector in the direction of given vector and then multiply it with given magnitude.

Let a⃗  = 2î – 3ĵ + 6k̂
Then, |a⃗ | = (2)2+(−3)2+(6)2−−−−−−−−−−−−−−−√
= 4+9+36−−−−−−−−√=49−−√ = 7 units

The unit vector in the direction of the given vector a⃗  is

Now, the vector of magnitude equal to 21 units
and in the direction of a is given by
21â = 21(27i^−37j^+67k^) = 6î – 9ĵ + 18k̂

Question 7.
Find a vector a of magnitude 5√2, making an angle of π4 with X-axis, π2 with Y-axis and an acute angle 0 with Z-axis. (All India 2014)
Answer:
Here, we have l = cos π4, m = cos π2 and n = cosθ
⇒ l = 12√, m = 0 and n = cosθ

Question 8.
Write a unit vector in the direction of the sum of the vectors a⃗  = 2î + 2ĵ – 5k̂ and b⃗  = -2î + ĵ – 7k̂. (Delhi2014C)
Answer:
113(4î + 3ĵ – 12k̂)

Question 9.
Find the value of p for which the vectors 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are parallel. (All India 2014)
Answer:
Given, 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are two parallel vectors, so their direction ratios will be proportional.

Question 10.
Write the value of cosine of the angle which the vector a⃗  = î + ĵ + k̂ makes with Y-axis. (Delhi 2014C)
Answer:
Given, a⃗  = î + ĵ + k̂
Now, unit vector in the direction of a⃗  is

∴ Cosine of the angle which given vector makes with Z-axis is 13√

Question 11.
Find the angle between X-axis and the vector î + ĵ + k̂. (All India 2014C)
Answer:
Let a⃗  = î + ĵ + k̂
Now, unit vector in the direction of a⃗  is

So, angle between X-axis and the vector
î + ĵ + k̂ is cos α = 13√ ⇒ α = cos-1 (13√)
[∵ â = lî + mĵ + nk̂ and cos α = l ⇒ α = cos-1l]

Question 12.
Write a vector in the direction of the vector î – 2ĵ + 2k̂ that has magnitude 9 units. (Delhi 2014C)
Answer:
3î – 6ĵ + 6k̂

Question 13.
Write a unit vector in the direction of vector PQ→, where P and Q are the points (1, 3, 0) and (4, 5, 6), respectively. (Foreign 2014)
Answer:
First, find the vector PQ−→− by using the formula (x2 – x1)î + (y2 – y1)ĵ + (z2 – z1)k̂, then required unit vector is given by PQ−→−|PQ−→−|

Given points are P (1, 3, 0) and Q (4, 5, 6).
Here, x1 = 1, y1 = 3, z1 = 0
and x2 = 4, y2 = 5, z2 = 6
So, vector PQ = (x2 – x1)k̂ + (y2 – y1)ĵ + (z2 – z1)k̂
= (4 – 1)î + (5 – 3)ĵ + (6 – 0)k̂
= 3î + 2ĵ + 6k̂
∴ Magnitude of given vector

Hence, the unit vector in the direction of PQ→ is

Question 14.
If a unit vector a⃗  makes angle π3 with î, π4 with ĵ and an acute angle θ with k̂, then find the value of θ. (Delhi 2013)
Answer:
Here, we have
l = cosπ3, m = cos π4 and n = cos θ

Question 15.
Write a unit vector in the direction of the sum of vectors a⃗  = 2î – ĵ + 2k̂ and b⃗  = – î + ĵ + 3k̂. (Delhi 2013)
Answer:
126√i^+526√k^

Question 16.
If a⃗  = xî +2ĵ – zk̂ and b⃗  = 3î – yĵ + k̂ are two equal vectors, then write the value of x + y + z. (Delhi 2013)
Answer:
Two vectors are equal, if coefficients of their components are equal.
Given, a⃗ =b⃗  ⇒ xî + 2ĵ – zk̂ = î – yĵ + k̂

On comparing the coefficient of components, we get
x = 3, y = -2, z = -1
Now, x + y + z = 3 – 2 – 1 = 0

Question 17.
P and Q are two points with position vectors 3a⃗  – 2b⃗  and a⃗  + b⃗ , respectively. Write the position vector of a point R which divides the line segment PQ in the ratio 2 : 1 externally. (All India 2013)
Answer:
−a⃗ +4b⃗ 

Question 18.
L and M are two points with position vectors 2a⃗  – b⃗  and a⃗  + 2b⃗ , respectively. Write the position vector of a point N which divides the line segment LM in the ratio 2 : 1 externally. (All India 2013)
Answer:
5b⃗ 

Question 19.
A and B are two points with position vectors 2a⃗  – 3b⃗  and 6b⃗  – a⃗ , respectively. Write the position vector of a point P which divides the line segment AB internally in the ratio 1:2. (All India 2013)
Answer:
Given, A and B are two points with position vectors 2a⃗  – 3b⃗  and 6b⃗  – a⃗ , respectively. Also, point P divides the line segment AB in the ratio 1 : 2 internally.

Question 20.
Find the sum of the vectors a⃗  = î – 2ĵ + k̂ b⃗  = – 2î + 4ĵ + 5k̂ and c⃗  = î – 6ĵ – 7k̂. (Delhi 2012)
Answer:
Given vectors are a⃗  = î – 2ĵ + k̂ b⃗  = – 2î + 4ĵ + 5k̂ and c⃗  = î – 6ĵ – 7k̂.
Sum of the vectors a⃗ , b⃗  and c⃗  is
a⃗ +b⃗ +c⃗  = (î – 2ĵ + k̂) + (- 2î + 4ĵ + 5k̂) + (î – 6ĵ – 7 k̂)
= – 4ĵ – k̂

Question 21.
Find the sum of the following vectors. a⃗  = î – 3k̂, b⃗  = 2ĵ – k̂, c⃗  = 2î – 3ĵ + 2k̂. (Delhi 2012)
Answer:
3î – ĵ – 2k̂

Question 22.
Find the sum of the following vectors. a⃗  = î – 2ĵ, b⃗  = 2î – 3 ĵ, c⃗  = 2î + 3k̂. (Deihi 2012)
Answer:
5î – 5ĵ + 3k̂

Question 23.
Find the scalar components of AB→ with initial point A (2,1) and terminal point B(- 5, 7). (All India 2012)
Answer:
Given initial point is A (2,1) and terminal point is B (- 5, 7), then scalar component of AB−→− are
x2 – x1 = – 5 – 2 = – 7and y2 – y1 = 7 – 1 = 6.

Question 24.
For what values of a⃗ , the vectors 2î – 3ĵ + 4k̂ and aî + 6ĵ – 8k̂ are collinear? (Delhi 2011)
Answer:
If a⃗  and b⃗  are collinear, then use the condition a⃗  = λb⃗ , where λ is some scalar.

Let given vectors are a⃗  = 2î – 3ĵ + 4k̂ and a⃗  = aî + 6ĵ – 8k̂
We know that, vectors a⃗  and b⃗  are said to be collinear, if
a⃗  = k. b⃗ , where k is a scalar.
∴ 2î – 3ĵ + 4k̂ = k(aî + 6ĵ – 8k̂)

On comparing the coefficients of î and ĵ, we get
2 = ka and -3 = 6k ⇒ k = –12
∴ 2 = –12a ⇒ a = -4

Question 25.
Write the direction cosines of vector -2î + ĵ – 5k̂. (Delhi 2011)
Answer:
Direction cosines of the vector aî + bĵ + ck̂ are

Question 26.
Write the position vector of mid-point of the vector joining points P(2, 3, 4) and Q (4, 1, – 2). (Foreign 2011)
Answer:
Mid-point of the position vectors
a⃗  = a1î + a2ĵ + a3k̂ and
b⃗  = b1î + b2ĵ + b3k̂ is a⃗ +b⃗ 2 or (a1+b1)i^+(a2+b2)j^+(a3+b3)k^2

Given points are P(2, 3, 4) and Q(4,1,-2) whose position vectors are OP−→− = 2 î + 5ĵ + 4k̂ and OQ−→− =4î + ĵ – 2k̂.
Now, position vector of mid-point of vector joining points P(2, 3, 4) and Q(4, 1, – 2) is

Question 27.
Write a unit vector in the direction of vector a⃗  = 2î + ĵ + 2k̂. (All India 2011; Delhi 2009)
Answer:
We know that, unit vector in the direction of â is â = a⃗ |a⃗ |

Required unit vector in the direction of vector
a⃗  = 2î + ĵ + 2k̂

Question 28.
Find the magnitude of the vector a⃗  = 3î – 2ĵ + 6k̂. (All India 2011C: Delhi 2008)
Answer:
Magnitude of a vector r = xî + yĵ + zk̂ is |r⃗ | = x2+y2+z2−−−−−−−−−−√

Given vector is a = 3i – 2/ + 6fc.
∴ Magnitude of a⃗  = |a⃗ |=(3)2+(−2)2+(6)2−−−−−−−−−−−−−−−√
= 9+4+36−−−−−−−−√=49−−√ = 7 units

Question 29.
Find a unit vector in the direction of vector a⃗  = 2î + 3ĵ + 6k̂. (All India 2011C)
Answer:
27i^+37j^+67k^

Question 30.
If A, B and C are the vertices of a ΔABC, then what is the value of AB−→−+BC−→−+CA−→− ? (Delhi 2011C)
Answer:
Let ΔABC be the given triangle.

Now, by triangle law of vector addition,

Question 31.
Find a unit vector in the direction of a⃗  = 2î – 3ĵ + 6k̂. (Delhi 2011c)
Answer:
27i^−37j^+67k^

Question 32.
Find a vector in the direction of a⃗  = 2î – ĵ + 2k̂, which has magnitude 6 units. (Delhi 2010C)
Answer:
4î – 2ĵ + 4k̂

Question 33.
Find the position vector of mid-point of the line segment AB, where A is point (3, 4, -2) and Bis point (1, 2, 4). (Delhi 2010)
Answer:
2î + 3ĵ + k̂

Question 34.
Write a vector of magnitude 9 units in the direction of vector -2î + ĵ + 2k̂. (All India 2010)
Answer:
-6î + 3ĵ + 6k̂

Question 35.
Write a vector of magnitude 15 units in the direction of vector î – 2ĵ + 2k̂. (Delhi 2010)
Answer:
5î – 10ĵ + 10k̂

Question 36.
What is the cosine of angle which the vector √2î + ĵ + k̂ makes with Y-axis? (Delhi 2010)
Answer:
12

Question 37.
Find a vector of magnitude 5 units and parallel to the resultant of a⃗  = 2î + 3ĵ – k̂ and b⃗  = î – 2ĵ + k̂. (Delhi 2011)
Answer:
First, find resultant of the vectors a and o, which is a⃗  + b⃗ . Then, find a unit vector in the direction of a⃗  + b⃗ . After this, the unit vector is multiplying by 5.

Given, a⃗  = 2î + 3ĵ – k̂ and b⃗  = î – 2ĵ + k̂.
Now, resultant of above vectors = a⃗  + b⃗ 
= (2î + 3ĵ – k̂) + (î – 2ĵ + k̂) = 3î + ĵ

Question 38.
Let a⃗  = î + ĵ + k̂, b⃗  = 4î – 2ĵ + 8k̂ and c⃗  = î – 2ĵ + k̂. Find a vector of magnitude 6 units, which is parallel to the vector 2a⃗  – b⃗  + 8 c⃗ . (All India 2010)
Answer:
First, find the vector 2 a – b + 3c, then find a unit vector in the direction of 2a-b + 3c.
After this, the unit vector is multiplying by 6.
Given, a⃗  = î + ĵ + k̂, b⃗  = 4î – 2ĵ + 8k̂ and c⃗  = î – 2ĵ + k̂
∴ 2a⃗ −b⃗ +3c⃗ 
= 2 (î + ĵ + k̂) – (4î – 2ĵ + 3k̂) + 3 (î – 2ĵ + k̂)
= 2î + 2ĵ + 2k̂ – 4î + 2ĵ – 3k̂ + 3î – 6ĵ + 3k̂
⇒ 2a⃗ −b⃗ +3c⃗  = î – 2ĵ + 2k̂

Now, a unit vector in the direction of vector

Hence, vector of magnitude 6 units parallel to the Vector 2a⃗ −b⃗ +3c⃗  = 6(13i^−23j^+23k^)
= 2î – 4ĵ + 4k̂

Question 39.
Find the position vector of a point R, which divides the line joining two points P and Q whose position vectors are 2a⃗  + b⃗  and a⃗  – 8b⃗  respectively, externally in the ratio 1 : 2. Also, show that P is the mid-point of line segment RO. (Delhi 2010)
Answer:
Given, OP−→− = Position vector of P = 2a⃗  + b⃗ 
and OQ−→− = Position vector of Q = a⃗  – 3b⃗ 

Let OR be the position vector of point R, which divides PQ in the ratio 1 : 2 externally

Now, we have to show that P is the mid-point of RQ,

Hence, P is the mid-point of line segment RQ.

Product of Two Vectors and Scalar Triple Product

Question 1.
Find the magnitude of each of the two vectors a⃗  and b⃗ , having the same magnitude such that the angle between them is 60° and their scalar product is 92. (CBSE 2018)
Answer:

Question 2.
Find the value of [î, k̂, ĵ], (CBSE 2018C)
Answer:
[î, k̂, ĵ] = î ∙ (k̂× ĵ)
= -[[î, k̂, ĵ] = –∣∣∣∣100010001∣∣∣∣ = – 1

Question 3.
Find λ and μ, if (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂) = 0. (All India 2016)
Answer:
Given, (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂)

= î(3μ + 9λ) ĵ k̂
On comparing the coefficients of î, ĵ and k̂ , we get
3μ + 9λ = 0, – μ + 27 = 0 and – λ – 9 = 0
⇒ μ = 27 and – λ = 9
⇒ μ = 27 and λ = – 9
Also, the values of μ and λ satisfy the equation
3μ + 9λ = 0.
Hence, μ = 27 and λ = – 9.

Question 4.
Write the number of vectors of unit length perpendicular to both the vectors a⃗  = 2î + ĵ + 2k̂ and b⃗  = ĵ + k̂. (All India 2016)
Answer:
We know that, unit vectors perpendicular to a⃗ 
and b⃗  are ±(a⃗ ×b⃗ |a⃗ ×b⃗ |)
So, there arc two unit vectors perpendicular to the given vectors.

Question 5.
If a⃗ ,b⃗ ,c⃗  are unit vectors such that a⃗ +b⃗ +c⃗ =0→ = 0, then write the value of a⃗ ⋅b⃗ +b⃗ ⋅c⃗ +c⃗ ⋅a⃗ . (Foreign 2016)
Answer:

Question 6.
If |a⃗ ×b⃗ |2+|a⃗ ⋅b⃗ |2 = 400 and |a⃗ | = 5, then write the value of |b⃗ |. (Foreign 2016)
Answer:

Question 7.
Find λ, if the vectors a⃗  = î + 3ĵ + k̂, b⃗  = 2 î – ĵ – k̂ and c⃗  = λĵ + 3 k̂ are coplanar. (Delhi 2015)
Answer:

⇒ 1(- 3 + λ) – 3(6) + 1(2λ) = 0
[expanding along R1]
⇒ – 3 + λ – 18 + 2λ = 0
⇒ 3λ = 21
∴ λ = 7

Question 8.
If a⃗  = 7î + ĵ – 4k̂ and b⃗  = 2î + 6ĵ + 3k̂, then find the projection of a⃗  on b⃗ . (Delhi 2015)
Answer:

Question 9.
If â, b̂ and ĉ are mutually perpendicular unit vectors, then find the value of |2â + b̂ + ĉ |. (All India 2015)
Answer:
Given â, b̂ and ĉ are mutually perpendicular unit vectors, i.e.

Question 10.
Write a unit vector perpendicular to both the vectors a⃗  = î + ĵ + k̂ and b⃗  = î + ĵ. (All India 2015)
Answer:
First, determine perpendicular vectors of a⃗  and b⃗ , i.e., a⃗ ×b⃗ . Further , determine perpendicular unit vector by using formula a⃗ ×b⃗ |a⃗ ×b⃗ |.

Given vector are a⃗  = î + ĵ + k̂ and b⃗  = î + ĵ
As we know the, vectors a⃗ ×b⃗  is perpendicular to both the vectors, so let us first evaluate a⃗ ×b⃗ .
Then, a⃗ ×b⃗  = ∣∣∣∣∣i^11j^11k^10∣∣∣∣∣
= î(0 -1) – ĵ(0 – 1) + k̂(1 – 1)
= – î + ĵ
Then , the unit vector perpendicular to both a⃗  and b⃗  is given by

Question 11.
Find the area of a parallelogram whose adjacent sides are represented by the vectors 2 î – 3 k̂ and 4 ĵ + 2 k̂. (Foreign 2015)
Answer:
Let adjacent sides of a parallelogram bc
a⃗  = 2 î – 3 k̂ and b⃗  = 4 ĵ + 2 k̂.

Question 12.
If a⃗  and b⃗  are perpendicular vectors, |a⃗  + b⃗ | = 13 and |a⃗ | = 5, then find the value of |b⃗ |. (All India 2014)
Answer:

Question 13.
If a⃗  and b⃗  are two unit vectors such that a⃗  + b⃗  is also a unit vector, then find the angle between a⃗  and b⃗ . (Delhi 2014)
Answer:

Question 14.
Find the projection of the vector î + 3ĵ + 7k̂ on the vector 2î – 3 ĵ + 6k̂. (Delhi 2014)
Answer:
let a⃗  = î + 3ĵ + 7k̂ and a⃗  = 2î – 3 ĵ + 6k̂

Question 15.
Write the projection of vector î + ĵ + k̂ along the vector ĵ. (Foreign 2014)
Answer:
1

Question 16.
Write the value of the following. î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ). (Foreign 2014)
Answer:
we have, î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ)
= î × ĵ + î × k̂ × ĵ × k̂ + ĵ × î + k̂ × î + k̂ × ĵ
[∵ cross product is distributive over addition]
= k̂ – ĵ + î – k̂ + ĵ – î = 0⃗ 
[∵ î × ĵ = k̂, î × k̂ = – ĵ, ĵ × k̂ = î, ĵ × î = – k̂, k̂ × î = ĵ, k̂ × ĵ = – î ]

Question 17.
If vectors a⃗  and b⃗  are such that |a⃗ | = 3, |b⃗ | = 2/3 and a⃗  × b⃗  is a unit vector, then write the angle between a⃗  and b⃗ . (Delhi 2014: All India 2010)
Answer:

Question 18.
Find a⃗ ⋅(b⃗ ×c⃗ ), if a⃗  = 2î + ĵ + 3k̂, b⃗  = -î + 2ĵ + k̂, and c⃗  = 3î + ĵ + 2k̂. (All India 2014)
Answer:
Given, a⃗  = 2î + ĵ + 3k̂, b⃗  = -î + 2ĵ + k̂, and c⃗  = 3î + ĵ + 2k̂.

= 2(4 – 1) – 1 (- 2 – 3) + 3( – 1 – 6)
= 2 × 3 – 1 × (-5) + 3 × (- 7)
= 6 + 5 – 21 = 11 – 21 = – 10

Question 19.
If a⃗  and b⃗  are unit vectors, then find the angle between a⃗  and b⃗ , given that (√3a⃗  – b⃗ ) is a unit vector. (Delhi 2014C)
Answer:

Question 20.
If |a⃗ | = 8, |b⃗ | = 3 and||a⃗ ×b⃗ || = 12, find the angle between a⃗  and b⃗ . (All India 2014C)
Answer:
let θ be the angle between latex]\vec{a}[/latex] and b⃗ .

Question 21.
Write the projection of the vector a⃗  = 2 î – ĵ + k̂ on the vector b⃗  = î + 2ĵ + 2k̂. (Delhi 2014 C)
Answer:
23

Question 22.
Write the value of λ, so that the vectors a = 2î + λĵ + k̂ and b = î – 2ĵ + 3k̂ are perpendicular to each other. (Delhi 2013C, 2008)
Answer:
Given vectors are a⃗  = 2î + λĵ + k̂
and b⃗  = î – 2ĵ + 3k̂
Since, vectors are perpendicular.
∴ a⃗ ⋅b⃗  = 0
⇒ (2î + λĵ + k̂) ∙ (î – 2ĵ + 3k̂)
⇒ 2 – 2λ + 3 = 0
∴ λ = 5/2

Question 23.
Write the projection of (b⃗  + c⃗ ) on a⃗ , where a⃗  = 2î – 2ĵ + k̂, b⃗  = î + 2ĵ – 2k̂ and c⃗  = 2î – ĵ + 4k̂. (All India 2013 C)
Answer:

Question 24.
Write the projection of the vector 7î + ĵ – 4k̂ on the vector 2î + 6 ĵ + 3k̂. (Delhi 2013C)
Answer:
87

Question 25.
If a⃗  and b⃗  are two vectors such that |a⃗  + b⃗ | = |a⃗ |, then prove that vector 2a⃗  + b⃗  is perpendicular to vector b. (Delhi 2013)
Answer:

Question 26.
Find |x⃗ |, if for â unit vector a, (x⃗ −a⃗ )⋅(x⃗ +a⃗ ) = 15. (All India 2013)
Answer:
Given, a⃗  is a unit vector. Then, |a⃗ | = 1.

Question 27.
Find λ, when projection of a⃗  = λî + ĵ + 4k̂ on b⃗  = 2 î + 6 ĵ + 3k̂ is 4 units. (Delhi 2012)
Answer:
Given, a⃗  = λî + ĵ + 4k̂ on b⃗  = 2 î + 6 ĵ + 3k̂ and projection of a⃗  and b⃗  = 4.

⇒ 2λ + 18 = 28
⇒ 2λ = 10
∴ λ = 5

Question 28.
Write the value of (k̂ × ĵ) . î + ĵ . k̂. (All India 2012)
Answer:
Use the results k̂ × ĵ = – î
ĵ ∙ k̂ and î ∙ î = 1 and simplify it.

Given, (k̂ × î) ∙ î + ĵ ∙ k̂ = (- î) ∙ î + ĵ ∙ k̂
= – (î ∙ î) + 0 = – 1 [∵ (î ∙ î) = 1]

Question 29.
If a⃗ ⋅a⃗  = 0 and a⃗ ⋅b⃗  = 0, then what can be concluded about the vector b⃗ ? (Foreign 2011)
Answer:

From Eqs. (i) and (ii). it may be concluded that b⃗  is either zero or non-zero perpendicular vector.

Question 30.
Write the projection of vector î – ĵ on the vector î + ĵ. (All India 2011)
Answer:
0

Question 31.
Write the angle between vectors a⃗  and b⃗  with magnitudes √3 and 2 respectively, having a⃗ . b⃗  = √6. (All India 2011)
Answer:
let θ be the angle between a⃗  and b⃗ , then use the following formula
cos θ = a⃗ ⋅b⃗ |a⃗ ||b⃗ |.

Question 32.
For what value of λ are the vectors î + 2λĵ + k̂ and 2î + ĵ – 3k̂ perpendicular? (All India 2011C)
Answer:
12

Question 33.
If |a⃗ | = √3, |b⃗ | = 2 and angle between a⃗  and b⃗  is 60°, then find a⃗ .b⃗ . (Delhi 2011C)
Answer:

Question 34.
Find the value of λ, if the vectors 2î + λĵ + 3k and 3î + 2ĵ – 4k̂ are perpendicular to each other. (All India 2010c)
Answer:
3

Question 35.
If |a⃗ | = 2, |b⃗ | = 3 and a⃗ .b⃗  = 3, then find the projection of b⃗  on a⃗ . (All India 2010C)
Answer:

Question 36.
If a⃗  and b⃗  are two vectors, such that |a⃗ ⋅b⃗ |=|a⃗ ×b⃗ |, then find the angle between a⃗  and b⃗ . (All India 2010)
Answer:
Use the following formulae:
a⃗ ⋅b⃗  = |a⃗ ||b⃗ | cos θ
and |a⃗ ×b⃗ | = |a⃗ ||b⃗ | sin θ
where, θ is the angle between a⃗  and b⃗ .

Question 37.
Find λ, if (2î + 6ĵ + 14k̂) × (î – λĵ + Ik̂) = 0. (All India 2010)
Answer:
– 3

Question 38.
If the sum of two unit vectors a and b is a unit vector, show that the magnitude of their difference is √3. (Delhi 2019, 2012c)
Answer:
let c⃗  = a⃗  + b⃗ . Then, according to given condition c⃗  is a unit vector, i.e. |c⃗ | = 1.

[taking positive square root, as magnitude cannot be negative]

Question 39.
If a⃗  = 2î + 5ĵ + k̂, b⃗  = î – 2 ĵ + k̂ and c⃗  = – 3î + ĵ + 2k̂, find [a⃗ b⃗ c⃗ ]. (Delhi 2019)
Answer:

= 2(- 4 – 1) – 3(2 + 3) + 1(1 – 6)
= – 10 – 15 – 5 = – 30

Question 40.
If |a⃗ | = 2, |b⃗ | = 7 and a⃗ ×b⃗  = 3î + 2ĵ + 6k̂, find the angle between a⃗  and b⃗ . (All India 2019)
Answer:
let θ be the angle between a⃗  and b⃗ .

Question 41.
Find the volume of cuboid whose edges are given by -5î + 7ĵ + 5k̂, -5î + 7ĵ – 5k̂ and 7î – 5 ĵ – 5k̂. (All India 2019)
Answer:

= |- 3 (- 21 – 15) – 7 (15 + 21) + 5(25 – 49)|
= |1108 – 252 – 120|
= 264 cubic units

Question 42.
Show that the points A(-2î + 5ĵ + 5k̂), B(î + 2 ĵ + 5k̂) and C(7î – k̂) are collinear. (All India 2019)
Answer:

Question 43.
Find |a⃗ ×b⃗ |, if a⃗  = 2î + ĵ + 5k̂ and b⃗  = 3î + 5ĵ – 2k̂. (All India 2019)
Answer:
We have, a⃗  = 2î + ĵ + 3k̂ and b⃗  = 3î + 5ĵ – 2k̂
∴ a⃗ ×b⃗  = ∣∣∣∣∣i^23j^15k^3−2∣∣∣∣∣
= î ( – 2 – 15) – ĵ (- 4 – 9) + k̂(10 – 3)
= – 17î + 13ĵ + 7k̂

Question 44.
If θ is the angle between two vectors î – 2 ĵ + 3k̂ and 3î – 2 ĵ + k̂, find sin θ. (CBSE 2018)
Answer:
let a⃗  = î – 2 ĵ + 3k̂ and b⃗  3î – 2 ĵ + k̂

Question 45.
If a⃗ +b⃗ +c⃗  = 0 and |a⃗ | = 5, |b⃗ | = 6 and |c⃗ | = 9, then find the angle between a⃗  and b⃗ . (CBSE 2018C)
Answer:

(5)2 + 2 × 5 × 6 × cos θ + (6)2 = (9)2
⇒ 25 + 60 cos θ + 36 = 81
⇒ cos θ = 2060=13
⇒ θ = cos-1(13)

Question 46.
If î + ĵ + k̂, 2î + 5ĵ, 5î + 2ĵ – 5k̂ and î – 6ĵ – k̂ respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether AB−→− and CD−→− are collinear or not. (Delhi 2019)
Answer:

Question 47.
The scalar product of the vector a⃗  = î + ĵ + k̂ with a unit vector along the sum of the vectors b⃗  = 2î + 4ĵ – 5k̂ and c⃗  = λî + 2ĵ + 5k̂ is equal to 1. Find the value of λ and hence find the unit vector along b⃗  + c⃗ . (All India 2019)
Answer:

Question 48.
Let a⃗  = 4 î + 5ĵ – k̂, b⃗  = î – 4ĵ + 5k̂ and c⃗  = 3î + ĵ – k̂. Find a vector which is perpendicular to both c⃗  and b⃗  and d⃗ ⋅a⃗  = 21. (CBSE 2018)
Answer:
We have, a⃗  = 4 î + 5ĵ – k̂, b⃗  = î – 4ĵ + 5k̂ and c⃗  = 3î + ĵ – k̂.
Since, d⃗  is perpendicular to both c⃗  and b⃗ .

= λ[î(5 – 4) – ĵ(15 + 1) + k̂(- 12 – 1)]
= λ(î – 16ĵ – 13k̂)
Also, it is given that a⃗ ⋅a⃗  = 21
∴ λ(î – 16ĵ – 13k̂) ∙ (4î + 5ĵ – k̂) = 21
⇒ λ(4 – 80 + 13) = 21
⇒ λ(- 63) = 21
⇒ λ = −13
Now from Eq. (j), we get
d⃗  = –−13(î – 16ĵ – 13k̂)

Question 49.
Find x such that the four points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D (7, 7, 2) are coplanar. (CBSE 2018C)
Answer:
Given points are A(4, 4, 4), B (5, x, 8), C(5, 4, 1) and D(7, 7, 2), then position vectors of A, B, C and D respectively, are

⇒ 1(0 + 9) – (x – 4) (- 2 + 9) + 4(3 – 0) = 0
⇒ 9 – (x – 4) (7) + 12 = 0
⇒ 9 – 7x + 28 + 12 = 0
⇒ 49 – 7x = 0
⇒ 7x = 49
⇒ x = 7

Question 50.
Find the value of x such that the points A(3, 2, 1), B(4, x, 5), C(4, 2,- 2) and D (6, 5, -1) are coplanar. (All India 2017)
Answer:
5

Question 51.
If a⃗ , b⃗  and c⃗  are three mutually perpendicular vectors of the same magnitude, then prove that a⃗ +b⃗ +c⃗  is equally inclined with the vectors a⃗ , b⃗  and c⃗ . (Delhi 2017, 2013C, 2011)
Answer:
If three vectors a⃗ , b⃗  and c⃗  are mutually perpendicular to each other, then a⃗ ⋅b⃗ =b⃗ ⋅c⃗  = c⃗ ⋅a⃗  = 0 and if all three vectors a⃗ , b⃗  and c⃗  are equally inclined with the vector (a⃗ +b⃗ +c⃗ ) that means each vector a⃗ , b⃗  and c⃗  makes equal angle with (a⃗ +b⃗ +c⃗ ) by using formula
cos θ = a⃗ ⋅b⃗ |a⃗ ||b⃗ |.

Question 52.
Using vectors, find the area of the ΔABC, whose vertices are A(1, 2, 5), 5(2, -1, 4) and C(4, 5, -1). (Delhi 2017; All India 2013)
Answer:
Let the position vectors of the verices A, B and C of ΔABC be

Question 53.
Let a⃗  = î + ĵ + k̂, b⃗  = î + 0 ∙ ĵ + 0 ∙ k̂ and c⃗  = c1î + c2ĵ + c3k̂, then
(a) Let c1 = 1 and c2 = 2, find c3 which makes a⃗ , b⃗  and c⃗  coplanar.
(b) If c2 = – 1 and c3 = 1, show that no value of c1 can make a⃗ , b⃗  and c⃗  coplanar. (Delhi 2017)
Answer:
Given, a⃗  = î + ĵ + k̂, b⃗  = î + 0 ∙ ĵ + 0 ∙ k̂ and c⃗  = c1î + c2ĵ + c3k̂
The given vectors are coplanar iff [a⃗ b⃗ c⃗ ] = 0

(a) If c1 = 1 and c2 = 2,
Then, from Eq.(i), we get
∣∣∣∣11110210c3∣∣∣∣ = 0

⇒ – 1(c3 – 0) + 1(2 – 0) = 0
⇒ – c3 + 2 = 0
⇒ – c3 = – 2
⇒ c3 = 2

(b) If c2 = – 1 and c3 = 1, then from Eq. (i), we get
∣∣∣∣11c110−1101∣∣∣∣ = 0
⇒ 1(0) – 1(1 – 0) + 1(- 1 – 0) = 0
⇒ 0 – 1 – 1 = 0
⇒ – 2 ≠ 0
∴ No value of c1 can make 1’ and coplanar.
Hence proved

Question 54.
Show that the points A, B, C with position vectors 2î – ĵ + k̂, î – 5ĵ – 5k̂ and 5î – 4ĵ – 4k̂ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle. (All India 2017)
Answer:

Question 55.
Show that the vectors a⃗ , b⃗  and c⃗  are coplanar, if a + b, 6+ c and c+ a are coplanar. (Delhi 2016, Foreign 2014)
Or
Prove that, for any three vectors a⃗ , b⃗  and c⃗ ,[a⃗ +b⃗ b⃗ +c⃗ c⃗ +a⃗ ]=2[a⃗ b⃗ c⃗ ]. (Delhi 2014)
Answer:

Question 56.
Show that the four points A (4, 5, 1), B(0, -1, -1), C(3, 9, 4) and D (-4, 4, 4) are coplanar. (All India 2016)
Or
Show that the four points A, B, C and D with position vectors 4î + 5ĵ + k̂, – ĵ – k̂, 3î + 9ĵ + 4k̂ and 4(- î + ĵ + k̂), respectively are coplanar. (All India 2014)
Answer:
Let the position vector of points A, B, C and D are

= – 4(12 + 3) + 6 (- 3 + 24) – 2(1 + 32)
= – 60 + 126 – 66 = 0
Hence, the four points A, B, C and D are coplanar.

Question 57.
The two adjacent sides of a parallelogram are 2î – 4ĵ – 5k̂ and 2î + 2 ĵ + 3k̂. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram. (All India 2016)
Answer:
Let ABCD be the given parallelogram with

Question 58.
If a⃗ ×b⃗ =c⃗ ×d⃗  and a⃗ ×c⃗ =b⃗ ×d⃗ , then show that a⃗ −d⃗  is parallel to b⃗ −c⃗ , where a⃗ ≠d⃗  and b⃗ ≠c⃗ . (Foreign 2016; Delhi 2009)
Answer:
Use the result, if two vectors are parallel, then their cross-product will be a zero vector.

Question 59.
If r⃗  = xî + yĵ + zk̂, find (r⃗ ×i^)⋅(r⃗ ×j^) + xy. (Delhi 2015)
Answer:

Question 60.
If a⃗  = î + 2ĵ + k̂, b⃗  = 2î + ĵ and c⃗  = 3î – 4 ĵ – 5k̂, then find a unit vector perpendicular to both of the vectors (a⃗ −b⃗ ) and (c⃗ −b⃗ ). (All India 2015)
Answer:

Question 61.
Find the value of λ so that the four points A, B,C and D with position vectors 4 î + 5ĵ + k̂, -ĵ – k̂,3i + Xj+4k and – 4 î + 4ĵ + 4 k̂, respectively are coplanar. (Delhi 2015C)
Answer:
Use the condition that four points with position vectors A⃗ ,B⃗ ,C⃗  and D⃗  are coplanar, if
[AB−→−,AC−→−,AD−→−]=0→ = 0.

On expanding along R1, we get
⇒ – 4(3λ – 15 + 3) + 6(- 3 + 24) – 2(1 + 8λ – 40) = 0
⇒ – 4(3λ – 12) + 6(21) – 2(8λ – 39) = 0
⇒ – 12λ + 48 + 126 – 16λ + 78 = 0
⇒ – 28λ + 252 = 0
λ = 9

Question 62.
Prove that [a⃗ b⃗ +c⃗ d⃗ ]=[a⃗ b⃗ d⃗ ]+[a⃗ c⃗ d⃗ ]. (All India 2015C)
Answer:

Question 63.
If a⃗  = 2î – 3ĵ + k̂, b⃗  = – î + k̂, c⃗  = 2 ĵ – k̂ are three vectors, find the area of the parallelogram having diagonals  and . (Delhi 2014C)
Answer:

Question 64.
Vectors a⃗ ,b⃗  and c⃗  are such that a⃗ +b⃗ +c⃗ =0→ and |a⃗ | = 3, |b⃗ | = 5 and |c⃗ | = 7. Find the angle between a⃗  and b⃗ . (Delhi 2014,2008; All India 2008)
Answer:
π3

Question 65.
The scalar product of the vector a⃗  = î + ĵ + k̂ with a unit vector along the sum of vectors b⃗  = 2î + 4ĵ – 5k̂ and c⃗  = λî + 2ĵ + 3k̂ is equal to one. Find the value of λ and hence, find the unit vector along b⃗  + c⃗ . (All India 2014)
Or
The scalar product of vector i + j + k with the unit vector along the sum of vectors 2î + 4ĵ – 5k̂ and λî + 2ĵ + 3k̂ is equal to one. Find the value of λ. (All India 2009,2008C)
Answer:
First, determine the unit vector of b⃗ +c⃗ , i.e. b⃗ +c⃗ |b⃗ +c⃗ |. Further put a⃗ ⋅(b⃗ +c⃗ )|b⃗ +c⃗ | = 1 and then determine the value of λ.

⇒ (λ + 6)2 = λ2 + 4λ + 44 [squaring both sides]
⇒ λ2 + 36 + 12λ + 4λ + 44
⇒ 8λ = 8
⇒ λ = 1
Hence, the value of λ is 1.
On substituting the value of λ in Eq. (1), we get Unit vector along b⃗ +c⃗ 

Question 66.
Find the vector p⃗  which is perpendicular to both α⃗  = 4î + 5ĵ – k̂ and β⃗  = î – 4ĵ + 5k̂ and p⃗ . q⃗  = 21, where q⃗  = 3i + j – k. (All India 2014C)
Answer:

= î (25 – 4) – ĵ (20 + 1) + k̂(- 16 – 5)
= î(21) – ĵ(21) + k̂(- 21)
= 21î – 21ĵ – 21k̂
So, p⃗  = 21λk̂—21λ?—21λk [fromEq.(i)] ….. (ii)
Also, given that p⃗ ⋅q⃗  = 21
∴ (21λî – 21λĵ – 21λk̂) . (3î + ĵ – k̂) = 21
⇒ 63λ – 21λ + 21λ = 21
⇒ 63λ = 21
⇒ λ = 13
On putting λ = 13 in Eq. (ii), we get

which is the required vector.

Question 67.
Find the unit vector perpendicular to both of the vectors a⃗ +b⃗  and a⃗ −b⃗  where, a⃗  = î + ĵ + k̂ and b⃗  = î + 2ĵ + 3k̂. (Foreign 2014)
Answer:

⇒ 2x + 3y + 4z = 0 …… (ii)
and (xî + yĵ + zk̂) . (- ĵ – 2k̂) = 0
⇒ – y – 2z = 0
⇒ y = – 2z
On putting the value of yin Eq. (ii), we get
2x + 3 (- 2z) + 4z = 0
⇒ x = z
On substituting the value of x and y in Eq. (1),
we get
⇒ z2 + 4z2 + z2 = 1
⇒ 6z2 = 1
⇒ z = ± 16√
then, x = ± 16√
and y = ± 26√
Hence, the required vectors are

Question 68.
Find the unit vector perpendicular to the plane ABC where the position vectors of A, B and C are 2î – ĵ + k̂, î + ĵ + 2k̂ and 2î + 3k̂, respectively. (All India 2014C)
Answer:
A unit vector perpendicular to plane ABC is
AB−→−×AC−→−|AB−→−×AC−→−|

Let O be the origin of reference.

Question 69.
Dot product of a vector with vectors î – ĵ + k̂, 2î + ĵ – 3k̂ and î + ĵ + k̂ are respectively 4, 0 and 2. Find the vector. (Delhi 2013C)
Answer:

⇒ a1 + a2 + a3 = 2
On subtracting Eq. (iii) from Eq. (i), we get
– 2a2 = 2
⇒ a2 = – 1
On substituting a2 = – 1 in Eq. (ii) and (iii),
we get
2a2 – 3a3 = 1 …… (iv)
⇒ a1 + a3 = 3
On multiplying Eq. (v) by 3 and then adding with Eq. (iv), we get
5a1 = 1 + 9 = 10
⇒ a1 = 2
On substituting a1 = 2 in Eq. (v), we get
a3 = 1
Hence, the vector is a⃗  = 2î – ĵ + k̂

Question 70.
Find the values of λ for which the angle between the vectors a⃗  = 2λ2î + 4λĵ + k̂ and b⃗  = 7î – 2ĵ + λk̂ is obtuse. (All India 2013C)
Answer:
let θ be the obtuse angle between the vectors

14λ2 – 7λ < 0
= 2λ2 – λ < 0
Either λ < 0, 2λ – 1 > 0 or λ > 0, 2λ – 1 < 0
= Either λ < 0, λ > 12 or λ > 0, λ < 12 Clearly, first option is impossible. ∴ λ > 0, λ < 12
0 < λ < 12
λ ∈ (0,12)

Question 71.
If a, b and c are three vectors such that each one is perpendicular to the vector obtained by sum of the other two and |a⃗ | = 3, |b⃗ | = 4 and |c⃗ | = 5, then prove that |a⃗ +b⃗ +c⃗ | = 5√2. (All India 2013C, 2010C)
Or
If a⃗ ,b⃗  and c⃗  are three vectors, such that |a⃗ | = 3, |b⃗ | = 4 and |c⃗ | = 5 and each one of these is perpendicular to the sum of other two, then find |a⃗ +b⃗ +c⃗ |. (All India 2011C, 2010C)
Answer:

Question 72.
If a⃗  = 3î – ĵ and b⃗  = 2î + ĵ – 3k̂, then express b⃗  in the form b⃗ =b⃗ 1+b⃗ 2, where b⃗ 1∥a⃗  and b⃗ 2⊥a⃗ . (All India 2013C)
Answer:
Given a⃗  = 3î – ĵ and b⃗  = 2î + ĵ – 3k̂
Let b1→ = x1î + y1ĵ + z1k̂ are two vectors such that b1→+b2→=b⃗ ,b1→∥a⃗  and b2→⊥a⃗ 

Consider, b⃗ 1+b⃗ 2=b⃗ 
⇒ (x1 + x2)î + (y1 + y2)ĵ + (z1 + z2)k̂ = 2î + ĵ – 3k̂

On comparing the coefficient of î ĵ and k̂ both sides, we get
x1 + x2 = 2
y1 + y2 = 1
z1 + z2 = -3

Now, consider b1→∥a⃗ 
⇒ x13=y1−1=z10
⇒ x1 = 3λ, y1 = -λ,and z1 = 0 …(iv)

On substituting the values of x, y and z, from Eq. (iv) to Eq. (i), (ii) and (iii), respectively, we get
x2 = 2- 3λ, y2 = -1 + λ and z2 = -3 …(v)
Since, b2 ± a , therefore b2 a = 0
⇒ 3x2 – y2 = 0
⇒ 3 (2 – 3λ) – (1 + λ) = 0
⇒ 6 – 9λ – 1 – λ = 0
⇒ 5 – 10λ = 0
⇒ λ = 12

On substituting λ = 12 in Eqs. (iv) and (v), we get

Question 73.
If a⃗  = î + ĵ + k̂ and b⃗  = ĵ – k̂, then find a vector c⃗ , such that a⃗ ×c⃗ =b⃗  and a⃗ ⋅c⃗  = 3. (Delhi 2013, 2008)
Answer:
Given a⃗  = î + ĵ + k̂ and b⃗  = ĵ – k̂
Let c⃗  = xî + yĵ + zk̂

= î (z – y) – ĵ(z – x) + k̂(y – x)
Now, a⃗ ×c⃗ =b⃗  [given]
= î(z – y) + ĵ(x – z) + k̂(y – x)
= 0î + 1ĵ + (-1)k̂ [∵ b⃗  = ĵ – k̂]

On comparing the coefficients from both sides, we get
z – y = 0,x – z = 1, y – x = -1
⇒ y = z and x – y = 1…(i)

Also given, a⃗ ⋅c⃗  = 3
⇒ (î + ĵ + k̂) . (xî + yĵ + zk̂) = 3
⇒ x + y + z = 3 (1)
⇒ x + 2y = 3 [∵ y = z] …(ii)

On subtracting Eq. (i) from Eq. (ii), we get
3y = 2
⇒ y = 23 = z [∵ y = z]

From Eq. (i),
x = 1 + y + 1 = 1 + 23=53
Hence, c⃗ =53i^+23j^+23k^

Question 74.
If a⃗  = î – ĵ + 7k̂ and b⃗  = 5î – ĵ + λk̂, then find the value of λ, so that  and  are perpendicular vectors. (All India 2013)
Answer:
Use the result that if a⃗  and b⃗  are perpendicular, then their dot product should be zero and simplify it.
Given, a⃗  = î – ĵ + 7k̂ and b⃗  = 5î – ĵ + λk̂
Then, a⃗ +b⃗  = (î – ĵ + 7k̂) + (5î – ĵ + λk̂)
= 6î – 2ĵ + (7 + λ) k̂
and a – = (î – ĵ + 7k̂) – (5î – j ̂+ λk̂)
= -4î + (7 – λ)k̂

Since, (a⃗ +b⃗ ) and (a⃗ −b⃗ ) are perpendicular
vectors, then (a⃗ +b⃗ )⋅(a⃗ −b⃗ ) = 0
⇒ [6î – 2ĵ + (7 + λ)k̂]- [-4î + (7 – λ)k̂] = 0 (1)
⇒ -24 + (7+ X)(7 – X) =0
⇒ 49 – λ2 = 24
⇒ λ2 = 25
∴ λ = ± 5

Question 75.
If p = 5î + λĵ – 3k̂ and q = î + 3ĵ – 5k̂, then find the value of λ, so that p⃗ +q⃗  and p⃗ −q⃗  are perpendicular vectors. (All India 2013)
Answer:
λ = ± 1

Question 76.
If a⃗ ,b⃗  and c⃗  are three vectors, such that |a⃗ | = 5, |b⃗ | = 12, |c⃗ | = 13 and a⃗ +b⃗ +c⃗  = 0, then find the value of a⃗ ⋅b⃗ +b⃗ ⋅c⃗ +c⃗ ⋅a⃗ . (Delhi 2012)
Answer:
-169

Question 77.
Let a⃗  = î + 4ĵ + 2k̂, b⃗  = 3î – 2ĵ + 7k̂ and c⃗  = 2î – ĵ + 4k̂. Find a vector p⃗ , which is perpendicular to both a⃗  and b⃗  and p⃗ .c⃗  = 18. (All India 2012,2010)
Answer:
Given vectors are a⃗  = î + 4ĵ + 2k̂,
b⃗  = 3î – 2ĵ + 7k̂
and c⃗  = 2î – ĵ + 4k̂

Let p⃗  = xî + yĵ + zk̂
We have, p⃗  is perpendicular to both a⃗  and b⃗ .
p⃗ ⋅a⃗  = 0
⇒ (xî + yĵ + zk̂) – (î + 4ĵ + 7k̂) = 0
⇒ x + 4y + 2z = 0 ………….(i)

and p⃗ ⋅b⃗  = 0
⇒ (xî + yĵ + zk̂) . (3î – 2ĵ + 7k̂) = 0
⇒ 3x – 2y + 7z = 0 …(ii)

Also, given ~p-~c =18 (1)
⇒ (xî + yĵ + zk̂) . (2î – ĵ + 4k̂) = 0
⇒ 2x – y + 4z = 18 …(iii)

On multiplying Eq. (i) by 3 and subtracting it from Eq. (ii), we get
– 14y + z = 0 ..(iv)

Now, multiplying Eq. (i) by 2 and subtracting it from Eq. (iii), we get
– 9y = 18
⇒ y = -2

On putting y = -2 in Eq. (iv), we get
-14 (-2) + z = 0
⇒ 28 + z = 0
⇒ z = -28

On putting y = -2 and z = -28 in Eq. (i), we get
x + 4 (-2) + 2 (-28) = 0
⇒ x – 8 – 56 = 0
⇒ x = 64

Hence, the required vector is
p⃗  = xî + yĵ + zk̂
i.e. p⃗  = 64î – 2ĵ – 28k̂

Question 78.
Find a unit vector perpendicular to each of the vectors a⃗ +b⃗  and a⃗ −b⃗ , where a = 3î + 2 ĵ + 2k̂ and b = î + 2ĵ – 2k̂. (Delhi 2011)
Answer:
23i^−23j^−13k^

Question 79.
If a and 6 are two vectors, such that |a⃗ | = 2, |b⃗ | = 1 and a⃗ .b⃗  = 1, then find (3a⃗ −5b⃗ )⋅(2a⃗ +7b⃗ ). (Delhi 2011)
Answer:

Question 80.
If vectors a⃗  = 2î + 2ĵ + 3k̂, b⃗  = -î + 2ĵ + k̂ and c⃗  = 3î + ĵ are such that a⃗ +λb⃗  is perpendicular to c⃗ , then find the value λ. (Foreign 2011; All India 2009C)
Answer:
Given, a⃗  = 2î + 2ĵ + 3k̂,
b⃗  = -î + 2ĵ + k̂
and c⃗  = 3î + ĵ

Also, a⃗  + λb⃗  is perpendicular to c⃗ .
∴ (a⃗  + λb⃗ ).c⃗  = 0 …(i) [∵ when [latex][/latex], then a⃗ ⋅b⃗  = 0]
Now, a⃗  + λb⃗  = (2î + 2ĵ + 3k̂) + λ (-î + 2ĵ + k̂)
⇒ a⃗  + λb⃗  = î(2 – λ) + ĵ(2 + 2λ) + k̂(3 + λ)
Then, from Bq. (i), we get
[î (2 – λ) + ĵ (2 + 2λ) + k̂(3 + λ)].[3î + ĵ] = 0
⇒ 3(2 – λ) + 1(2+ 2k) = 0
⇒ 8 – λ = 0
∴ λ = 8

Question 81.
Using vectors, find the area of triangle with vertices A (1, 1, 2), 5(2, 3, 5) and C (1, 5, 5). (All India 2011)
Answer:
1261−−√ sq.units

Question 82.
Using vectors, find the area of triangle with vertices A (2, 3, 5), B (3, 5, 8) and C(2, 7, 8). (Delhi 2010C)
Answer:
1261−−√ sq.units

Some Extra Question 

Q. No. 2: Find the unit vector in the direction of the sum of the vectors

$\begin{array}{l}\overrightarrow{�}=2\widehat{�}-\widehat{�}+2\widehat{�}\\ \overrightarrow{�}=-\widehat{�}+\widehat{�}+3\widehat{�}\end{array}$

Solution:

$\begin{array}{l}\text{Let}\overrightarrow{�}\text{be the sum of}\overrightarrow{�}\text{and}\overrightarrow{�}.\end{array}$

The unit vector is:

Q. No. 3: Find the vector joining the points P(2, 3, 0) and Q(– 1, – 2, – 4) directed from P to Q.

Solution:

Since the vector is to be directed from P to Q, clearly P is the initial point and Q is the terminal point.

P(2, 3, 0) = (x1, y1, z1)

Q(-1, -2, -4) = (x2, y2, z2)

Vector joining the points P and Q is:

$\begin{array}{l}\text{Q. No. 4: Find a vector in the direction of a vector}5\widehat{�}-\widehat{�}+2\widehat{�}\text{which has a magnitude of 8 units.}\end{array}$

Solution:

$\begin{array}{l}���\overrightarrow{�}=5\widehat{�}-\widehat{�}+2\widehat{�}\end{array}$

$\begin{array}{l}\text{Hence, the vector in the direction of vector}5\widehat{�}-\widehat{�}+2\widehat{�}\text{which has a magnitude of 8 units is given by}\end{array}$

$\begin{array}{l}\text{Q. No. 5: Show that the vector}\widehat{�}+\widehat{�}+\widehat{�}\text{is equally inclined to the axes OX, OY and OZ.}\end{array}$

Solution:

$\begin{array}{l}���\overrightarrow{�}=\widehat{�}+\widehat{�}+\widehat{�}\end{array}$

$\begin{array}{l}\text{Therefore, the direction cosines of}\overrightarrow{�}���(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}).\end{array}$

$\begin{array}{l}\text{Let}�,�,�\text{be the angles formed by}\overrightarrow{�}\text{with the positive directions of x, y, and z-axes.}\end{array}$

Then,

cos α = 1/√3, cos β = 1/√3 cos γ = 1/√3

Hence, the given vector is equally inclined to axes OX, OY and OZ.

Q. No. 6: Show that the points A, B and C with position vectors

$\begin{array}{l}\overrightarrow{�}=3\widehat{�}-4\widehat{�}-4\widehat{�},\\ \overrightarrow{�}=2\widehat{�}-\widehat{�}+\widehat{�}\end{array}$

$\begin{array}{l}\overrightarrow{�}=\widehat{�}-3\widehat{�}-5\widehat{�}\end{array}$

form the vertices of a right-angled triangle.

Solution:

Position vectors of points A, B and C are respectively given as below.

$\begin{array}{l}\overrightarrow{�}=3\widehat{�}-4\widehat{�}-4\widehat{�},\\ \overrightarrow{�}=2\widehat{�}-\widehat{�}+\widehat{�}\end{array}$

$\begin{array}{l}\overrightarrow{�}=\widehat{�}-3\widehat{�}-5\widehat{�}\end{array}$

Therefore, ABC is a right-angled triangle.

$\begin{array}{l}\text{Q. No. 7: Find a vector}\overrightarrow{�}\text{of magnitude}3\sqrt{2}\text{units which makes an angle of}�/4,�/2\text{with y and z-axes, respectively.}\end{array}$

Solution:

From the given,

m = cos π/4 = 1/√2

n = cos π/2 = 0

Therefore, l2 + m2 + n2 = 1

l2 + (½) + 0 = 1

l2 = 1 – ½

l = ±1/√2

Hence, the required vector is:

Q. No. 8: Evaluate the product.

$\begin{array}{l}(3\overrightarrow{�}-5\overrightarrow{�}).(2\overrightarrow{�}+7\overrightarrow{�})\end{array}$

Solution:

Q. No. 9: Find all vectors of magnitude 10√3 that are perpendicular to the plane of

$\begin{array}{l}\widehat{�}+2\widehat{�}+\widehat{�}\end{array}$

$\begin{array}{l}-\widehat{�}+3\widehat{�}+4\widehat{�}\end{array}$

Solution:

$\begin{array}{l}���\overrightarrow{�}=\widehat{�}+2\widehat{�}+\widehat{�}\\ \overrightarrow{�}=-\widehat{�}+3\widehat{�}+4\widehat{�}\end{array}$

$\begin{array}{l}\text{Hence, the unit vector perpendicular to the plane of}\overrightarrow{�}\text{and}\overrightarrow{�}\text{is:}\end{array}$

$\begin{array}{l}\text{Therefore, the vectors of magnitude}10\sqrt{3}\text{that are perpendicular to the plane of}\overrightarrow{�}\text{and}\overrightarrow{�}\text{are:}\end{array}$

Q. No. 10: Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Solution:

Vertices of a triangle ABC are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Let AB and BC be the adjacent sides of triangle ABC.

Hence, the area of triangle ABC is √61/2 sq.units

Practice Questions For Class 12 Maths Chapter 10 Vector Algebra

1. Using vectors find the area of triangle PQR, whose vertices are P(1, 2, 3), Q(2, -1, 4) and C(4, 5, -1).
2. Use vector method to show that P,Q,R are Collinear – P(3, -5, 1), Q(-1, 0, 8) and C(7, -10, -6)
3. Find the angles between the lines whose direction ratios are 3, 2, -6 and 1, 2, 2. Find the angles of a triangle PQR whose vertices are P(-1, 3, 2), B(2, 3, 5) and C(3, 5, -2).
4. Prove that the points P(-2, 4, 7), Q (3, -6, -8) and R(1,-2,-2) are collinear.
5. Find the angle between the lines whose direction ratios are: 2, -3, 4 and 1, 2, 1.
6. Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.
7. $\begin{array}{l}\text{If A, B, C, D are the points with position vectors}\widehat{�}+\widehat{�}–\widehat{�},2\widehat{�}-\widehat{�}+3\widehat{�},2\widehat{�}-3\widehat{�},3\widehat{�}-2\widehat{�}+\widehat{�}\text{respectively. Find the projection of}\overrightarrow{��}\text{along}\overrightarrow{��}.\end{array}$
8. Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
9. Show that:
   
10. Show that each of the given three vectors is a unit vector: