Free Objective Test 01 Practice Test - 11th and 12th

Question 1

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 =

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

$I f \to a = (3, - 2, 1), \to b = (- 1, 1, 1)$ 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}})

SOLUTION

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

Question 3

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

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

A. a=b

B. a=c

C. c=a+b

D. a =b=c

SOLUTION

Solution : 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 7

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

\frac{π}{6}

\frac{π}{4}

\frac{π}{3}

\frac{π}{2}

SOLUTION

Solution : 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:

(0, \frac{1}{2})

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

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

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

SOLUTION

Solution : 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 9

A unit vector perpendicular to the plane of $a = 2^i - 6^j - 3^k, b = 4^i + 3^j -^k$ is

\frac{4^i + 3^j -^k}{\sqrt{26}}

\frac{2^i - 6^j - 3^k}{7}

\frac{3^i - 2^j + 6^k}{7}

\frac{2^i - 3^j - 6^k}{7}

SOLUTION

Solution : 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 10

If a is any vector then $(a \times^i)^{2} + (a \times^j)^{2} + (a \times^k)^{2}$ =

a^{2}

2 a^{2}

3 a^{2}

4 a^{2}

SOLUTION

Solution : B

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

Question 11

The area of the parallelogram whose diagonals are $^i - 3^j + 2^k, -^i + 2^j$ is

4 \sqrt{29} s q

.units

\frac{1}{2} \sqrt{21} s q

.units

10 \sqrt{3} s q

.units

\frac{1}{2} \sqrt{270} s q

.units

SOLUTION

Solution : 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 12

The volume of the parallellopiped whose coterminal edges are $2^i - 3^j + 4^k,^i + 2^j - 2^k, 3^i -^j +^k$ is

A. 5

B. 6

C. 7

D. 8

SOLUTION

Solution : C

Volume =| $∣ ∣ ∣ ∣ \begin{matrix} 2 & - 3 & 4 1 & 2 & - 2 3 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ | = | 2 (2 - 2) + 3 (1 + 6) + 4 (- 1 - 6) | = | 0 + 21 - 28 | = 7$ cubic units

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

{⃗ a}^{2} = {⃗ b}^{2} + {⃗ c}^{2}

⃗ a + ⃗ b = ⃗ c

N o n e o f t h e s e

SOLUTION

Solution : A

The position vector of the centroid of the triangle is $\frac{⃗ a + ⃗ b + ⃗ c}{3}$
Since, the triangle is an equilateral, therefore the orthocenter coincides with the centroid and hence $\frac{⃗ a + ⃗ b + ⃗ c}{3} = ⃗ 0 \Rightarrow ⃗ a + ⃗ b + ⃗ c = 0$

Question 14

If $\to a =^i +^j +^k, \to b = 2^i -^j +^k$ and $\to c =^i + x^j + y^k$ , are linearly dependent and $| \to c | = \sqrt{3}$ then $(x, y)$ is

(1, 1)

(- 2, 0)

(\frac{1}{5}, \frac{7}{5})

(- \frac{7}{5}, \frac{3}{5})

SOLUTION

Solution : A

Given that the three vectors are linearly dependent so
$\to c = l \to a + m \to b$
$\Rightarrow l + 2 m = 1$
$l - m = x$
$\Rightarrow x = 3 y - 2$
$l + m = y$
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 15

If $\to a^{'} =^i +^j, \to b^{'} =^i +^j + 2^k$ and $\to c^{'} = 2^i +^j -^k$ . Then altitude of the parallelopiped formed by the vectors $\to a, \to b, \to c$ having base formed by $\to b$ and $\to c$ is $(\to a, \to b, \to c$ and ${\to a}^{'}, {\to b}^{'}, {\to c}^{'}$ are reciprocal system of vectors)

1

\frac{3 \sqrt{2}}{2}

\frac{1}{\sqrt{6}}

\frac{1}{\sqrt{2}}

SOLUTION

Solution : D

Volume of the parallelepiped formed by $\to a^{'}, \to b^{'}, \to c^{'}$ is 4
$∴$ Volume of the parallelepiped formed by $\to a, \to b, \to c$ is $\frac{1}{4}$
$\to b \times \to c = \frac{1}{4} \to a^{'} ∴ ∣ ∣ \to b \times \to c ∣ ∣ = \frac{\sqrt{2}}{4} = \frac{1}{2 \sqrt{2}}$
$∴$ length of altitude = $\frac{1}{4} \times 2 \sqrt{2} = \frac{1}{\sqrt{2}}$ .

Free Objective Test 01 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

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