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

12th Grade > Mathematics

VECTOR ALGEBRA MCQs

Total Questions : 60 | Page 2 of 6 pages

Question 11. The vector

a^i + b^j + c^k

is a bisector of the angle between the vectors

^i +^j

and

^j +^k

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 12. If a is any vector then

(a \times^i)^{2} + (a \times^j)^{2} + (a \times^k)^{2}

Discuss Question

Answer: Option B. -> 2a2
:
B

(a \times^i)^{2} + (a \times^j)^{2} + (a \times^k)^{2} = 2 a^{2}

Question 13. If a, b, c are the position vectors of the vertices of an equilateral triangle whose orthocenter is at the origin, then

→a+→b+→c=→0
→a2=→b2+→c2
→a+→b=→c
None of these

Discuss Question

Answer: Option A. -> →a+→b+→c=→0
:
A
The position vector of the centroid of the triangle is

\frac{\to a + \to b + \to c}{3}

Since, the triangle is an equilateral, therefore the orthocenter coincides with the centroid and hence

\frac{\to a + \to b + \to c}{3} = \to 0 \Rightarrow \to a + \to b + \to c = 0

Question 14. A unit vector perpendicular to the plane of

a = 2^i - 6^j - 3^k, b = 4^i + 3^j -^k

4^i+3^j−^k√26
2^i−6^j−3^k7
3^i−2^j+6^k7
2^i−3^j−6^k7

Discuss Question

Answer: Option C. -> 3^i−2^j+6^k7
:
C

a \times b = ∣ ∣∣∣ ∣ \begin{matrix} ^i & ^j & ^k 2 & - 6 & - 3 4 & 3 & - 1 \end{matrix} ∣ ∣∣∣ ∣ =^i (6 + 9) -^j (- 2 + 12) +^k (6 + 24) = 15^i - 10^j + 30^k

| a \times b | = \sqrt{225 + 100 + 900} = 35

Unit vector normal to the plane =

\frac{15^i - 10^j + 30^k}{35} = \frac{3^i - 2^j + 6^k}{7}

Question 15. The area of the parallelogram whose diagonals are

^i - 3^j + 2^k, -^i + 2^j

4√29sq.units
12√21sq.units
10√3sq.units
12√270sq.units

Discuss Question

Answer: Option B. -> 12√21sq.units
:
B
Vector area =

\frac{1}{2} (a \times b) = \frac{1}{2} ∣ ∣∣∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 3 & 2 1 & 2 & 0 \end{matrix} ∣ ∣∣∣ ∣ = \frac{1}{2} [(0 - 4) -^j (0 + 2) +^k (2 - 3)] = \frac{1}{2} (- 4^i - 2^j -^k)

Area =

\frac{1}{2} \sqrt{16 + 4 + 1} = \frac{1}{2} \sqrt{21} s q .

units

Question 16. 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:

(0,12)
(12,1)
(1,32)
(12,32)

Discuss Question

Answer: Option A. -> (0,12)
:
A

90^{\circ} < θ < 180^{\circ} \Rightarrow a . b < 0 \Rightarrow (2 x^{2}^i + 4 x^j +^k) . (7^i - 2^j + x^k) < 0 \Rightarrow 14 x^{2} - 8 x + x < 0 \Rightarrow 14 x^{2} - 7 x < 0 \Rightarrow 7 x (2 x - 1) < 0 \Rightarrow 0 < x < \frac{1}{2}

Question 17. If

⃗ a^{'} =^i +^j, ⃗ b^{'} =^i +^j + 2^k

and

⃗ c^{'} = 2^i +^j -^k

. Then altitude of the parallelopiped formed by the vectors

⃗ a, ⃗ b, ⃗ c

having base formed by

⃗ b

and

⃗ c

(⃗ a, ⃗ b, ⃗ c

and

{⃗ a}^{'}, {⃗ b}^{'}, {⃗ c}^{'}

are reciprocal system of vectors)

1
3√22
1√6
1√2

Discuss Question

Answer: Option D. -> 1√2
:
D
Volume of the parallelepiped formed by

⃗ a^{'}, ⃗ b^{'}, ⃗ c^{'}

is 4

∴

Volume of the parallelepiped formed by

⃗ a, ⃗ b, ⃗ c

\frac{1}{4}

⃗ b \times ⃗ c = \frac{1}{4} ⃗ a^{'} ∴ ∣ ∣ ⃗ b \times ⃗ c ∣ ∣ = \frac{\sqrt{2}}{4} = \frac{1}{2 \sqrt{2}}

∴

length of altitude =

\frac{1}{4} \times 2 \sqrt{2} = \frac{1}{\sqrt{2}}

Question 18. If

⃗ a =^i +^j +^k, ⃗ b = 2^i +^j +^k

and

⃗ c =^i + x^j + y^k

, are linearly dependent and

| ⃗ c | = \sqrt{3}

then (x,y) is

(1, 1)
(-2, 0)
(15,75)
(−75,35)

Discuss Question

Answer: Option A. -> (1, 1)
:
A
Given that the three vectors are linearly dependent so

⃗ c = l ⃗ a + m ⃗ b

\Rightarrow l + 2 m = 1

l - m = x

\Rightarrow x = 3 y - 2

I +m =y
Also,

x^{2} + y^{2} + 1 = 3

10 y^{2} - 12 y + 2 = 0

\Rightarrow y = 1, \frac{1}{5}

x = 1, - \frac{7}{5}

Question 19. If |a+b| = |a-b| then (a,b) =

Discuss Question

Answer: Option D. -> π2
:
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 20. 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.

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