Vector Algebra(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

VECTOR ALGEBRA MCQs

Total Questions : 30 | Page 1 of 3 pages

If the points A,B and C have position vectors (2,1,1), (6,-1,2) and (14,-5,P) respectively and if they are collinear, then P =

Discuss Question

Answer: Option B. -> 4
:
B

$⃗ O A = 2^i +^j +^k, ⃗ O B = 6^i -^j + 2^k, ⃗ O C = 14^i - 5^j + P^k \Rightarrow ⃗ A B = ⃗ O B - ⃗ O A = 4^i - 2^j +^k, ⃗ A C = ⃗ O C - ⃗ O A = 12^i - 6^j + (p - 1)^k$
A, B, C are collinear $\Rightarrow ⃗ A C = λ ⃗ A B \Rightarrow 12^i - 6^j + (p - 1)^k = λ (4^i - 2^j +^k)$
$\Rightarrow λ = 3, p - 1 = 3 \Rightarrow p = 4.$

Question 2.

$I f \to a = (3, - 2, 1), \to b = (- 1, 1, 1)$ then the unit vector parallel to the vector $\to a + \to b$ is

$(\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})$
$(\frac{2}{5}, - \frac{1}{5}, \frac{2}{5})$
$(\frac{2}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}})$
$(- \frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}}, - \frac{2}{\sqrt{3}})$

Discuss Question

Answer: Option A. ->

(\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})

:
A

\to a + \to b = (3, - 2, 1) + (- 1, 1, 1) = (2, - 1, 2)

\Rightarrow ∣ ∣ ∣ \to a + \to b ∣ ∣ ∣ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

Unit vector parallel to

\to a + \to b

\pm \frac{a + b}{∣ ∣ ∣ \to a + \to b ∣ ∣ ∣} = \pm \frac{(2, - 1, 2)}{3} = (\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})

Question 3.

The points $2^i -^j -^k,^i +^j +^k, 2^i + 2^j +^k, 2^j + 5^k$ are

collinear
coplanar but not collinear
noncoplanar
none

Discuss Question

Answer: Option B. -> coplanar but not collinear
:
B

⃗ A B = -^i + 2^k, ⃗ A C =^j + 2^k, ⃗ A D = - 2^i +^j + 6^k,

A, B, C, D are not collinear.

B o x = ∣ ∣∣ ∣ \begin{matrix} - 1 & 0 & 2 0 & 1 & 2 - 2 & 1 & 6 \end{matrix} ∣ ∣∣ ∣ = - 1 (6 - 2) + 2 (0 + 2) = - 4 + 4 = 0.

∴

A, B, C, D are coplannar.

Question 4.

The vectors 3a-2b-4c, -a+2c, -2a+b+3c are

linearly dependent
linearly independent
collinear
none

Discuss Question

Answer: Option A. -> linearly dependent
:
A

B o x = ∣ ∣∣ ∣ \begin{matrix} 3 & - 2 & - 4 - 1 & 0 & 2 - 2 & 1 & 3 \end{matrix} ∣ ∣∣ ∣ = 3 (0 - 2) + 2 (- 3 + 4) - 4 (- 1 - 0) = - 6 + 2 + 4 = 0

∴

Given vectors are coplanar Given vectors are linarly dependent.

Question 5.

The vector $a^i + b^j + c^k$ is a bisector of the angle between the vectors $^i +^j$ and $^j +^k$ if

a=b
a=c
c=a+b
a =b=c

Discuss Question

Answer: Option B. -> a=c
:
B

a^i + b^j + c^k

=Angle bisector of

^i +^j

and

^j +^k = m

[\frac{^i +^j +^j +^k}{\sqrt{2}}] \Rightarrow a = \frac{m}{\sqrt{2}}, b = \sqrt{2} . m, c = \frac{m}{\sqrt{2}}

Question 6.

The ratio in which $^i + 2^j + 3^k$ divides the join of $- 2^i + 3^j + 5^k$ and $7^i -^k$ is

-3 : 2
1 : 2
2 : 3
-4 : 3

Discuss Question

Answer: Option B. -> 1 : 2
:
B
Ratio =-2-1: 1-7 =-3:-6=1:2

Question 7.

If |a+b| = |a-b| then (a,b) =

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{2}$

Discuss Question

Answer: Option D. ->

\frac{π}{2}

:
D

| a + b | = | a - b | \Rightarrow | a + b |^{2} = | a - b |^{2} \Rightarrow (a + b)^{2} = (a - b)^{2} \Rightarrow a^{2} + b^{2} + 2 a . b = a^{2} + b^{2} - 2 a . b \Rightarrow 4 a . b = 0 \Rightarrow a . b = 0 \Rightarrow (a, b) = 90^{\circ}

Question 8.

If the angle θ between the vectors $a = 2 x^{2}^i + 4 x^j +^k$ and $b = 7^i - 2^j + x^k$ is such that $90^{\circ}$ < $θ$ < $180^{\circ}$
then x lies in the interval: