Free Vector Algebra 03 Practice Test - 12th Grade - Commerce

Question 1

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

A. 2

B. 4

C. 6

D. 8

SOLUTION

Solution : 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

$I f ⃗ a =< 3, - 2, 1 > ⃗ b =< - 1, 1, 1 >$ then the unit vector parallel to the vector $⃗ a + ⃗ b$ 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}} >

SOLUTION

Solution : A

$⃗ a + ⃗ b =< 3, - 2, 1 > + < - 1, 1, 1 >=< 2, - 1, 2 >\Rightarrow | ⃗ a + ⃗ b | = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
Unit vector parallel to $⃗ a + ⃗ b$ is $\pm \frac{a + b}{| ⃗ a + ⃗ b |} = \pm \frac{< 2, - 1, 2 >}{3}$

Question 3

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

A. collinear

B. coplanar but not collinear

C. noncoplanar

D. none

SOLUTION

Solution : 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

A. linearly dependent

B. linearly independent

C. collinear

D. none

SOLUTION

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

A. -3 : 2

B. 1 : 2

C. 2 : 3

D. -4 : 3

SOLUTION

Solution : B

Ratio =-2-1: 1-7 =-3:-6=1:2

Question 6

The value of b such that the scalar product of the vector $^i +^j +^k$ with the unit vector parallel to the sum of the vectors $2^i + 4^j + 5^k$ and $b^i + 2^j + 3^k$ is one, is

A. -2

B. -1

C. 0

D. 1

SOLUTION

Solution : D

The unit vector parallel to the sum of the vectors $2^i + 4^j - 5^k$ and $b^i + 2^j + 3^k$ is
$^n = \frac{(2 + b)^i + 6^j - 2 ¨ k}{\sqrt{(2 + b)^{2}} + 6^{2} + (- 2)^{2}} = \frac{(2 + b)^i + 6^j - 2 ¨ k}{\sqrt{b^{2} + 4 b + 44}}$
Now, $(^i +^j +^k)$ . $^n$ =1
$\Rightarrow 2 + b + 6 - 2 = \sqrt{b^{2} + 4 b + 44} \Rightarrow b = 1$

Question 7

The vector $\to C$ , directed along the internal bisector of the angle between the vectors $\to a = 7^i - 4^j - 4^k$ and $\to b = - 2^i -^j + 2^k$ with $| \to c | = 5 \sqrt{6}$ , is

\pm (\frac{5}{3} (^i - 7^j + 2^k)

\frac{5}{3} (5^i - 5^j + 2^k)

\frac{5}{3} (^i - 7^j + 2^k)

\frac{5}{3} (- 5^i - 5^j + 2^k)

SOLUTION

Solution : A

The required vector $\to C$ is given by
$→C=λ(^a+^b)=λ(→a|→a|+→b|→b|)=λ{19(7^i−4^j−4^k)+13(−2^i−^j+2^k)}$
$\Rightarrow \to c = \frac{λ}{9} (^i - 7^j + 2^k) \Rightarrow | \to c | = \pm \frac{λ}{9} \sqrt{1 + 49 + 4} = \pm \frac{λ}{9} \sqrt{54}$
But $| \to c | = 5 \sqrt{6}$ (given)
$∴ \pm \frac{λ}{9} \sqrt{54} = 5 \sqrt{6} \Rightarrow λ = \pm 15$
Hence, $\to c = \pm \frac{15}{9} (^i - 7^j + 2^k) = \pm \frac{5}{3} (^i - 7^j + 2^k)$

Question 8

If the vectors $⃗ a = (c l o g_{2} x)^i - 6^j + 2^k$ and $⃗ b = (l o g_{2} x)^i + 2^j + 3 (c l o g_{2} x)^k$ make an obtuse angle for any $x \in (0, \infty)$ then c belongs to

(- \infty, 0)

(- \infty, - \frac{4}{3})

(- \frac{4}{3}, 0)

(- \frac{4}{3}, \infty)

SOLUTION

Solution : C

For the vectors $⃗ a$ and $⃗ b$ to be inclined at an obtuse angle, we must have
$⃗ a . ⃗ b < 0$ for all $x \in (0, \infty)$
$\Rightarrow c (l o g_{2} x)^{2} - 12 + 6 c (l o g_{2} x) < 0$ for all $x \in (0, \infty)$
$\Rightarrow c y^{2} + 6 c y - 12 < 0$ for all $y \in R$ , where $y = l o g_{2} x$
$\Rightarrow c < 0$ and $\Rightarrow 36 c^{2} + 48 c < 0 \Rightarrow c < 0$ and $c (3 c + 4) < 0$
$\Rightarrow c < 0$ and $- \frac{4}{3} < c < 0$
$\Rightarrow c \in (- \frac{4}{3}, 0)$

Question 9

The vectors $⃗ a = x^i + (x + 1)^j + (x + 2)^k, ⃗ b = (x + 3)^i + (x + 4)^j + (x + 5)^k$ and $⃗ c = (x + 6)^i + (x + 7)^j + (x + 8)^k$ are coplanar for

A. all values of x

B. x < 0

C. x > 0

D. None of these

SOLUTION

Solution : A

$⃗ a, ⃗ b, ⃗ c$ are coplanar, iff $[⃗ a ⃗ b ⃗ c] = 0$
We have, $[⃗ a ⃗ b ⃗ c] = ∣ ∣ ∣ ∣ \begin{matrix} x & x + 1 & x + 2 x + 3 & x + 4 & x + 5 x + 6 & x + 7 & x + 8 \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} x & x + 1 & x + 2 3 & 3 & 3 6 & 6 & 6 \end{matrix} ∣ ∣ ∣ ∣ [\begin{matrix} A p p l y i n g R_{2} \to & R_{2} - & R_{1}, R_{3} \to & R_{3} - & R_{1} \end{matrix}]$
= 0 for all x [ $∵ R_{1}$ and $R_{2}$ are proportional]

Question 10

If $⃗ a = 2^i + 3^j +^k, ⃗ b = 2^i + p^j + 3^k$ and $⃗ c = 2^i + 17^j + 3^k$ are coplanar vectors, then the value of p is

A. -4

B. -1

C. 4

D. -2

SOLUTION

Solution : A

Since, $⃗ a = 2^i + 3^j +^k, ⃗ b = 2^i + p^j + 3^k$ and $⃗ c = 2^i + 17^j + 3^k$ are coplanar,
therefore $[⃗ a ⃗ b ⃗ c] = ∣ ∣ ∣ ∣ \begin{matrix} 2 & 3 & 1 2 & p & 3 2 & 17 & - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow p = - 4$

Free Vector Algebra 03 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Related ExamsMatrices 03Definite Integrals and Areas 03Relations and Functions 03

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Matrices 03
Definite Integrals and Areas 03
Relations and Functions 03