Free Objective Test 02 Practice Test - 11th and 12th

Question 1

The three points whose position vectors are $^i + 2^j + 3^k, 3^i + 4^j + 7^k a n d - 3^i - 2^j - 5^k$

A. form the vertices of an equilateral triangle

B. form the vertices of a right angled triangle

C. are collinear

D. form the vertices of an isosceles triangle.

SOLUTION

Solution : C

If A, B, C are the given points respectively, then
$\to O A =^i + 2^j + 3^k, \to O B = 3^i + 4^j + 7^k, \to O C = - 3^i - 2^j - 5^k, \to A B = \to O B - \to O A = 2^i + 2^j + 4^k, \to A C = \to O C - \to O A = - 4^i - 4^j - 8^k = - 2 \to A B$
$∴ \to A B, \to A C$ are collinear $\Rightarrow A, B, C$ are collinear.

Question 2

If a,b represent $- - \to A B, - - \to B C$ respectively of a regular hexagon ABCDEF then $- - \to C D, - - \to D E, - - \to E F, - - \to F A$ are

A. b-a, -a, -b, a-b

B. A-b, a, b, b-a

C. b-a, a, b, a-b

D. A-b,-a,-b, b-a

SOLUTION

Solution : A

ABCDEF is a regular hexagon
$\Rightarrow ⃗ A D = 2 ⃗ B C, ⃗ E D = ⃗ A B, ⃗ F E = ⃗ B C, ⃗ F A = ⃗ D C$
Given $⃗ A B = a, ⃗ B C = b$
Now $⃗ A B + ⃗ B C + ⃗ C D = ⃗ A D \Rightarrow a + b + ⃗ C D = 2 ⃗ B C$
$\Rightarrow ⃗ C D = 2 b - (a + b) = b - a ⃗ D E = ⃗ B A = - ⃗ A B = - a, ⃗ E F = ⃗ C B = - ⃗ B C = - b ⃗ F A = ⃗ D C = - ⃗ C D = - (b - a) = a - b$

Question 3

If A=(1,3,-5) and B=(3,5,-3), then the vector equation of the plane passing through the midpoint of AB and perpendicular to AB is

r . (^i +^j +^k) = 2

r . (^i +^j -^k) = 2

r . (^i -^j +^k) = 2

N o n e

SOLUTION

Solution : A

$- - \to A B = - - \to O B - - - \to O A = (3^i + 5^j - 3^k) - (^i + 3^j - 5^k) = 2^i + 2^j + 2^k$
Midpoint of AB is (2, 4, -4)
Vector equation of the plane is $[r - (2^i + 4^j - 4^k)] . (2^i + 2^j + 2^k) = 0$
$\Rightarrow r . (^i +^j +^k) = 2 + 4 - 4 \Rightarrow r . (^i +^j +^k) = 2$

Question 4

A unit vector perpendicular to the plane determined by the points P(1,-1,2), Q(2,0,-1) and R(0,2,1) is

\frac{2^i +^j +^k}{\sqrt{6}}

\frac{2^i +^j +^k}{3}

\frac{2^i -^j -^k}{\sqrt{3}}

\frac{2^i -^j -^k}{3}

SOLUTION

Solution : A

$- - \to O P =^i -^j + 2^k, - - \to O Q = 2^i -^k, - - \to O R = 2^j +^k \Rightarrow - - \to P Q = - - \to O Q - - - \to O P =^i +^j - 3^k, - - \to P R = - - \to O R - - - \to O P = -^i + 3^j -^k$
$- - \to P Q \times - - \to P R = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 1 & - 3 - 1 & 3 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 8^i + 4^j + 4^k; | - - \to P Q \times - - \to P R | = \sqrt{64 + 16 + 16} = \sqrt{96} = 4 \sqrt{6}$
Required unit vectors = $\pm \frac{8^i + 4^j + 4^k}{4 \sqrt{6}} = \pm \frac{2^i +^j +^k}{\sqrt{6}}$

Question 5

If $x . a = 0, x \times b = c \times b$ then x =

c - \frac{c . a}{b . a} b

c - \frac{c . a}{b . a} a

a - \frac{c . a}{b . a} b

b - \frac{c . a}{b . a} b

SOLUTION

Solution : A

$x \times b = c \times b \Rightarrow (x - c) \times b = 0 \Rightarrow x - c$ is parallel to b
$\Rightarrow x - c = λ b$ for some scalar $λ \Rightarrow x = c + λ b$
$x . a = 0 \Rightarrow (c + λ b) . a = 0 \Rightarrow c . a + λ b . a = 0 \Rightarrow λ = - \frac{c . a}{b . a} \Rightarrow x = c - \frac{c . a}{b . a} b$

Question 6

If the vector a is perpendicular to b and c, |a|=2, |b|=3, |c|=4 and the angle between b and c is $\frac{2 π}{3}$ then |[a b c ]| =

A. 24

B. 12

12 \sqrt{3}

24 \sqrt{3}

SOLUTION

Solution : C

$| [a b c] | = | a . (b \times c) | = | a | | b \times c | | c o s (a, b \times c) | = | a | | b \times c | = | a | | b | c | s i n (b, c)$
$= 2.3.4 s i n (\frac{2 π}{3}) = 24. \frac{\sqrt{3}}{2} = 12 \sqrt{3}$ .

Question 7

The volume of the tetrahedron with vertices at (1,2,3), (4,3,2), (5,2,7), (6,4,8) is

\frac{22}{3}

\frac{11}{3}

\frac{1}{3}

\frac{16}{3}

SOLUTION

Solution : D

$[- - \to A B - - \to A C - - \to A D] = ∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 & - 1 4 & 0 & 4 5 & 2 & 5 \end{matrix} ∣ ∣ ∣ ∣ = 3 (0 - 8) - 1 (20 - 20) - 1 (8 - 0) = - 24 - 0 - 8 = - 32$
Volume of the tetrahedron = $\frac{1}{6} (32) = \frac{16}{3}$ cubic unit.

Question 8

If a,b,c are linearly independent, then $\frac{[2 a + b, 2 b + c, 2 c + a]}{[a, b, c]} =$

A. 9

B. 8

C. 7

D. None

SOLUTION

Solution : A

$\frac{[2 a + b, 2 b + c, 2 c + a]}{[a, b, c]} = ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 & 0 0 & 2 & 1 1 & 0 & 2 \end{matrix} ∣ ∣ ∣ ∣ = 2 (4 - 0) - 1 (0 - 1) = 8 + 1 = 9$

Question 9

Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then $- - \to O A + - - \to O B + - - \to O C + - - \to O D$ equals

- - \to O A

2 - - \to O P

3 - - \to O P

4 - - \to O P

SOLUTION

Solution : D

Since, the diagonals of a parallelogram bisect each other. Therefore, P is the middle point of AC and BD both.
$∴ - - \to O A + - - \to O C = 2 - - \to O P a n d - - \to O B + - - \to O D = 2 - - \to O P \Rightarrow - - \to O A + - - \to O B + - - \to O C + - - \to O D = 4 - - \to O P$

Question 10

If $\to a =^i + 2^j + 2^k$ and $\to b = 3^i + 6^j + 2^k$ , then the vector in the direction of $\to a$ and having magnitude as $| \to b |$ , is

7 (^i + 2^j + 2^k)

\frac{7}{9} (^i + 2^j + 2^k)

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

N o n e o f t h e s e

SOLUTION

Solution : C

The required vectors
$= | \to b |^a = \frac{| \to b |}{| \to a |} \to a = \frac{7}{3} (^i + 2^j + 2^k)$

Question 11

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 12

The vector $⃗ C$ , directed along the internal bisector of the angle between the vectors $⃗ a = 7^i - 4^j - 4^k$ and $⃗ b = - 2^i -^j + 2^k$ with $| ⃗ c | = 5 \sqrt{6}$ , 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 $⃗ C$ is given by
$⃗C=λ(^a+^b)=λ(⃗a|⃗a|+⃗b|⃗b|)=λ{19(7^i−4^j−4^k)+13(−2^i−^j+2^k)}$
$\Rightarrow ⃗ c = \frac{λ}{9} (^i - 7^j + 2^k) \Rightarrow | ⃗ c | = \pm \frac{λ}{9} \sqrt{1 + 49 + 4} = \pm \frac{λ}{9} \sqrt{54}$
But $| ⃗ c | = 5 \sqrt{6}$ (given)
$∴ \pm \frac{λ}{9} \sqrt{54} = 5 \sqrt{6} \Rightarrow λ = \pm 15$
Hence, $⃗ c = \pm \frac{15}{9} (^i - 7^j + 2^k) = \pm \frac{5}{3} (^i - 7^j + 2^k)$

Question 13

If the vectors $\to a = (c l o g_{2} x)^i - 6^j + 2^k$ and $\to b = (l o g_{2} x)^i + 2^j + 3 (c l o g_{2} x)^k$ 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 $\to a$ and $\to b$ to be inclined at an obtuse angle, we must have
$\to a . \to b < 0$ 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 14

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

A. all values of x

B. x < 0

C. x > 0

D. None of these

SOLUTION

Solution : A

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

Question 15

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 Objective Test 02 Practice Test - 11th and 12th

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

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsObjective Test 02Objective Test 03Objective Test 03

Related Exams
Objective Test 02
Objective Test 03
Objective Test 03