Projection of a vector on $$3$$ coordinate axes are $$6, - 3, 2$$ respectively. Then DC's of vector are-

$$\dfrac{6}{5},\dfrac{{ - 3}}{5},\dfrac{2}{5}$$

$$\dfrac{-6}{7},\dfrac{{ - 3}}{2},\dfrac{2}{7}$$

If the points $$\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$$ are collinear, then

$$k$$ has infinite number of real values

The Cartesian equation of line $$6x - 2 = 3y + 1 = 2z - 2$$ is given by

$$\dfrac{{3x - 1}}{6} = \dfrac{{3y - 1}}{3} = \dfrac{{z - 1}}{3}$$

The vector equation of line passing through the point $$(-1,-1,2)$$ and parallel to the line $$2x-2=3y+1=6z-2$$

$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +6\hat { k } )$$

$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (\hat { i } +2\hat { j } +3\hat { k } )$$

$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +\hat { k } )$$

If $$A=(1,2,-1), B=(2,0,3), C=(3,-1,2)$$ then the angle between $$\overline { AB } $$ and $$\overline { AC } $$ is

$$\cos ^{ -1 }{ \left( \cfrac { 15 }{ \sqrt { 21 } \sqrt { 11 } } \right) } $$

The direction cosines to two lines at right angles are (1,2,3) and (-2,$$\frac{1}{2}$$,$$\frac{1}{3}$$), then the direction cosine perpendicular to both given lines are:

$$\sqrt{\frac{25}{2198}}$$,$$\sqrt{\frac{19}{2198}}$$,$$\sqrt{\frac{729}{2198}}$$

$$\sqrt{\frac{24}{2198}}$$,$$\sqrt{\frac{38}{2198}}$$,$$\sqrt{\frac{730}{2198}}$$

$$\frac{1}{3}$$,-2,$$\frac{-7}{2}$$

MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 8

ABC is a triangle where

A = (2, 3, 5), B = (- 1, 2, 2)

$A = ( 2,3,5 ) , B = ( - 1,2,2 )$ and

C (λ, 5, μ)

$C (\lambda,5 , \mu )$ if the median through A is equally inclined to the positive axis then

λ + μ

$\lambda + \mu$ is

0%
7
0%
6
0%
15
0%
9

Explanation

A D

$AD$ is median

D (\frac{λ - 1}{2}, \frac{7}{2}, \frac{μ + 2}{2})

$D\left( {\frac{{\lambda - 1}}{2},\frac{7}{2},\frac{{\mu + 2}}{2}} \right)$

- - \to A D = (\frac{λ - 1}{2} - 2, \frac{7}{2} - 3, \frac{μ + 2}{2} - 5)

$\overrightarrow {AD} = \left( {\frac{{\lambda - 1}}{2} - 2,\frac{7}{2} - 3,\frac{{\mu + 2}}{2} - 5} \right)$

- - \to A D

$\overrightarrow {AD}$ is equally in lined.

∴

$\therefore$ paralle to

(1, 1, 1)

$(1,1,1)$

\begin{matrix} \frac{λ - 5}{2} = \frac{1}{2} = \frac{μ + 8}{2} λ = 6 μ = 9 λ + μ = 15 \end{matrix}

$\begin{array}{l} \frac { { \lambda -5 } }{ 2 } =\frac { 1 }{ 2 } =\frac { { \mu +8 } }{ 2 } \\ \lambda =6 \\ \mu =9 \\ \lambda +\mu =15 \end{array}$

Then,

Option

C

$C$ is correct answer.

Projection of a vector on

3

$3$ coordinate axes are

6, - 3, 2

$6, - 3, 2$ respectively. Then DC's of vector are-

0%
$6, - 3, 2$
0%
$\dfrac{6}{5},\dfrac{{ - 3}}{5},\dfrac{2}{5}$
0%
$\dfrac{-6}{7},\dfrac{{ - 3}}{2},\dfrac{2}{7}$
0%
$\dfrac{ 6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$

Explanation

Projection of a vector on coordinate axis are

X_{2} - X_{1}, Y_{2} - Y_{1}, Z_{2} - Z_{1}

$X_2-X_1, Y_2-Y_1, Z_2-Z_1$

X_{2} - X_{1} = 6, Y_{2} - Y_{1} = - 3, Z_{2} - Z_{1} = 2

$X_2-X_1=6, Y_2-Y_1=-3, Z_2-Z_1=2$

\sqrt{{(X_{2} - X_{1})}^{2} + {(Y_{2} - Y_{1})}^{2} + {(Z_{2} - Z_{1})}^{2}} = \sqrt{36 + 9 + 4} = 7

$\sqrt {{{\left( {{X_2} - {X_1}} \right)}^2} + {{\left( {{Y_2} - {Y_1}} \right)}^2} + {{\left( {{Z_2} - {Z_1}} \right)}^2}} = \sqrt {36 + 9 + 4} = 7$

The Direction cosines of the vector are

\frac{6}{7}, \frac{- 3}{7}, \frac{2}{7}

$\dfrac{6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$

If the foot of the perpendicular from

(0, 0, 0)

$(0,0,0)$ to a plane is

P (1, 2, 2)

$P(1,2,2)$ . Then, the equation of the plane is

0%
$-x+2y+8z-9=0$
0%
$x+2y+2z-9=0$
0%
$x+y+z-5=0$
0%
$x+2y-3z+3=0$

Explanation

Equation of normal

\Rightarrow^i + 2^j + 2^k

$\Rightarrow \ \hat i+2\hat j+2\hat k$

equation of plane

\Rightarrow

$\Rightarrow$

(x - 1) 1 + (x y - 2) 2 + (z - 2) 2 = 0

$(x-1)1+(xy-2)2+(z-2)2=0$

x - 1 + 2 y - 4 + 2 z - 4 = 0

$x-1+2y-4+2z-4=0$

x + 2 y + 2 z - 9 = 0

$x+2y+2z-9=0$

P (1, 1, 1)

$P\left(1,1,1\right)$ and

Q (λ, λ, λ)

$Q\left(\lambda,\lambda,\lambda\right)$ are two points in the space such that

P Q = \sqrt{27},

$PQ=\sqrt{27},$ the value of

λ

$\lambda$ can be

0%
$-4$
0%
$-2$
0%
$2$
0%
$0$

Explanation

{P Q}^{2} = {(λ - 1)}^{2} + {(λ - 1)}^{2} + {(λ - 1)}^{2}

${PQ}^{2}={\left(\lambda-1\right)}^{2}+{\left(\lambda-1\right)}^{2}+{\left(\lambda-1\right)}^{2}$

\Rightarrow 3 {(λ - 1)}^{2} = 27

$\Rightarrow\,3{\left(\lambda-1\right)}^{2}=27$

\Rightarrow {(λ - 1)}^{2} = 9

$\Rightarrow\,{\left(\lambda-1\right)}^{2}=9$

\Rightarrow λ - 1 = \pm 3

$\Rightarrow\,\lambda-1=\pm\,3$

\Rightarrow λ = 1 \pm 3

$\Rightarrow\,\lambda=1\pm\,3$

\Rightarrow λ = 1 + 3,

$\Rightarrow\,\lambda=1+3,$ or

1 - 3

$\,1-3$

∴ λ = 4,

$\therefore\,\lambda=4,$ or

- 2

$-2$

A (2, 3, 7), B (- 1, 3, 2)

$A(2,3,7),B(-1,3,2)$ and

C (q, 5, r)

$C(q,5,r)$ are the vertices of

Δ A B C

$\Delta ABC$ . If the median through A is equally inclined to the coordinate axes then the coordinates of the vertex C is

0%
$(7,5,14)$
0%
$(7,5,12)$
0%
$(7,5,10)$
0%
$(7,5,16)$

If the points

¯ a + ¯ b, ¯ a - ¯ b, ¯ a + k ¯ b

$\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$ are collinear, then

0%
$k$ has only one real value
0%
$k$ has two real value
0%
$k$ has no real values
0%
$k$ has infinite number of real values

Explanation

As the

3

$3$ point should be collinear area of triangle termed by then should be zero

considering

2

$2$ direction to be

(¯ a + ¯ b) - (¯ a - ¯ b) = 2 ¯ b

$(\bar{a}+\bar{b})-(\bar{a}-\bar{b})=2\bar{b}$

(¯ a + ¯ b) - (¯ a + k ¯ b) = (1 - k) ¯ b

$(\bar{a}+\bar{b})-(\bar{a}+k\bar{b})=(1-k)\bar{b}$

(2 ¯ b) \times (1 - k) ¯ b = 0

$(2\bar{b})\times (1-k)\bar{b}=0$

this will be for any value of

k

$k$ as cross produced of

2

$2$ linear vector

= 0

$=0$

A line

A B

$AB$ in three-dimensional space males angles

45^{o}

${45}^{o}$ and

120^{o}

${120}^{o}$ with the positive x-axis and the positive y-axis respectively. If

A B

$AB$ makes an acute angle

θ

$\theta$ with the positive z-axis, then

θ

$\theta$ equal

0%
${45}^{o}$
0%
${60}^{o}$
0%
${75}^{o}$
0%
${30}^{o}$

Explanation

'l', 'm', 'n' be the Direction cosine of the line

l = c o s α = c o s 45^{\circ} = \frac{1}{\sqrt{2}}

$l = cos \alpha = cos45^{\circ} = \dfrac{1}{\sqrt{2}}$

m = c o s β = c o s 120^{\circ} = \frac{- 1}{2}

$m = cos\beta = cos120^{\circ} = \dfrac{-1}{2}$

n = c o s θ

$n = cos\theta$

we know

l^{2} + m^{2} + n^{2} = 1

$l^{2}+m^{2}+n^{2} = 1$

\Rightarrow d \frac{1}{2} + \frac{1}{4} + c o s^{2} θ = 1

$\Rightarrow d\dfrac{1}{2}+\dfrac{1}{4}+cos^{2}\theta = 1$

\Rightarrow c o s^{2} θ = 1 - \frac{1}{2} - \frac{1}{4}

$\Rightarrow cos^{2}\theta = 1-\dfrac{1}{2}- \dfrac{1}{4}$

\Rightarrow c o s^{2} θ = \frac{1}{4}

$\Rightarrow cos^{2}\theta = \dfrac{1}{4}$

\Rightarrow c o s = \frac{1}{2}

$\Rightarrow cos = \dfrac{1}{2}$

∴ θ = 60^{\circ}

$\therefore \theta = 60^{\circ}$

1072324_1183560_ans_ebea3676557a4eaab7a1b39b82e92fe1.png

If the points

(α, - 1), (2, 1)

$(\alpha, - 1), (2, 1)$ and

(4, 5)

$(4, 5)$ are collinear, then find

α

$\alpha$ by vector method.

0%
$4$
0%
$1$
0%
$8$
0%
None of these

Explanation

If there points are collinear then vectors from one to another will have scalar triple produced

0

$0$ .Point

(α, - 1), (2, 1), (4, 5)

$\left(\alpha,-1\right), \left(2,1\right), \left(4,5\right)$

(2 - α)^i + 2^j - ¯ A

$\left( 2-\alpha \right) \hat { i } +2\hat { j } -\bar { A }$

2^i + 4^j - ¯ B

$2\hat { i } +4\hat { j } -\bar { B }$

(4 - α)^i + 6^j - ¯ C

$\left( 4-\alpha \right) \hat { i } +6\hat { j } -\bar { C }$

¯ A . (¯ B \times ¯ C) = 0

$\bar { A } .\left( \bar { B } \times \bar { C } \right) =0$

((2 - α)^i + 2^j) (2^i + 4^j) \times ((4 - α)^i + 6^j) ((2 - α)^i + 2^j) . [12^k - 16^k + 4 α^k] = 0

$\left( \left( 2-\alpha \right) \hat { i } +2\hat { j } \right) \left( 2\hat { i } +4\hat { j } \right) \times \left( \left( 4-\alpha \right) \hat { i } +6\hat { j } \right) \\ \left( \left( 2-\alpha \right) \hat { i } +2\hat { j } \right) .\left[ 12\hat { k } -16\hat { k } +4\alpha \hat { k } \right] =0$

4 α = 4

$4\alpha =4$

α = 1

$\alpha =1$

Also the direction vector will be proportion

(2 - α, 2) = λ (4 - 2.5 - 1)

$\left( 2-\alpha,2 \right)=\lambda\left( 4-2.5-1\right)$

(2 - α, 2) = λ (2, 4)

$\left( 2-\alpha,2 \right)=\lambda\left( 2,4\right)$

λ = \frac{1}{2}

$\lambda=\dfrac{1}{2}$ as

2 = 4 λ

$2=4\lambda$

2 - α = 1

$2-\alpha=1$

∴ α = 1

$\therefore \alpha=1$

The Cartesian equation of line

6 x - 2 = 3 y + 1 = 2 z - 2

$6x - 2 = 3y + 1 = 2z - 2$ is given by

0%
$\dfrac{{3x - 1}}{3} = \dfrac{{3y + 1}}{6} = \dfrac{{z - 1}}{3}$
0%
$\dfrac{{3x + 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$
0%
$\dfrac{{3x - 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$
0%
$\dfrac{{3x - 1}}{6} = \dfrac{{3y - 1}}{3} = \dfrac{{z - 1}}{3}$

Explanation

Solution -

6 x - 2 = 3 y + 1 = 2 z - 2

$6x-2 = 3y+1 = 2z-2$

\frac{6 x - 2}{6} = \frac{3 y + 1}{6} = \frac{2 z - 2}{6}

$\dfrac{6x-2}{6} = \dfrac{3y+1}{6} = \dfrac{2z-2}{6}$

\frac{x - \frac{1}{3}}{1} = \frac{y + \frac{1}{3}}{2} = \frac{z - 1}{3}

$\dfrac{x-\dfrac{1}{3}}{1} = \dfrac{y+\dfrac{1}{3}}{2} = \dfrac{z-1}{3}$

(1, 2, 3)

$(1,2,3)$

Point passing through line

(1 / 3, - 1 / 3, 1)

$(1/3,-1/3,1)$

\frac{3 x - 1}{3} = \frac{3 y + 1}{6} = \frac{z - 1}{3}

$\dfrac{3x-1}{3} = \dfrac{3y+1}{6} = \dfrac{z-1}{3}$

A is correct.

The direction ratios of the joining

A (1, 2, 1)

$A(1,\,2,\ 1)$ and

(2, 1, 2)

$(2,\ 1,\ 2)$ are

0%
$3,\ 3,\ 3$
0%
$-1,\ 1,\ -1$
0%
$3,\ 1,\ 3$
0%
$\dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}}$

Explanation

We have given points are:-

A (1, 2, 1)

$A\left( {1,2,1} \right)$

B (2, 1, 2)

$B\left( {2,1,2} \right)$

Now,

- - \to B A = (1 - 2) ˆ i + (2 - 1) ˆ j + (1 - 2) ˆ k

$\overrightarrow {BA} = \left( {1 - 2} \right)\widehat i + \left( {2 - 1} \right)\widehat j + \left( {1 - 2} \right)\widehat k$

= - ˆ i + ˆ j - ˆ k

$= - \widehat i + \widehat j - \widehat k$

direction ratio are

a = - 1

$a=-1$

b = 1

$b=1$

c = - 1

$c=-1$

Hence,

option

B

$B$ is correct.

The direction ratios of

A B

$AB$ are

- 2, 2, 1

$- 2, 2, 1$ . If coordinates of A are

(4, 1, 5)

$( 4,1,5 )$ and

l (A B) = 6

$l( A B ) = 6$ , then coordinates of

B

$B$ ?

0%
$( 0,5 , - 7 )$
0%
$( 8 , - 3,3 )$
0%
$( 0,7,5 )$
0%
$( 8,3,3 )$

If the lines

L_{1} a n d L_{2}

${L}_{1}\ and {L}_{2}$ are given by

¯ r = (¯ i + 2 ¯ j - ¯ k) + t (¯ 2 i - 3 ¯ j + ¯ k) a n d ¯ r = (¯ i + ¯ j + ¯ k) + s (2 ¯ i + ¯ j - ¯ k)

$\bar { r } =\left( \bar { i } +2\bar { j } -\bar { k } \right) +t\left( \bar { 2i } -3\bar { j } +\bar { k } \right) \ and\bar { r } =\left( \bar { i } +\bar { j } +\bar { k } \right) +s\left( 2\bar { i } +\bar { j } -\bar { k } \right)$ , then

0%
${L}_{1}\ and {L}_{2}$ are perpendicular
0%
${L}_{1}\ and {L}_{2}$ are parallel
0%
$\left ({L}_{1},{L}_{2}\right)={45}^{o}$
0%
$\left ({L}_{1},{L}_{2}\right)={60}^{o}$

The vector equation of line passing through the point

(- 1, - 1, 2)

$(-1,-1,2)$ and parallel to the line

2 x - 2 = 3 y + 1 = 6 z - 2

$2x-2=3y+1=6z-2$

0%
$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (3\hat { i } +2\hat { j } +\hat { k } )$
0%
$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +6\hat { k } )$
0%
$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (\hat { i } +2\hat { j } +3\hat { k } )$
0%
$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +\hat { k } )$

Explanation

Given

L_{1} : 2 x - 2 = 3 y + 1 = 6 z - 2

$L_1 : 2x - 2 = 3y + 1 = 6z - 2$

\Rightarrow 2 (x - 1) = 3 (y + \frac{1}{3}) = 6 (z - \frac{1}{3})

$\Rightarrow 2 (x - 1) = 3 ( y + \dfrac{1}{3}) = 6(z - \dfrac{1}{3})$

\Rightarrow \frac{x - 1}{3} = \frac{y + \frac{1}{3}}{2} = \frac{z - \frac{1}{3}}{1}

$\Rightarrow \dfrac{x - 1}{3} = \dfrac{y + \dfrac{1}{3}}{2} = \dfrac{z - \dfrac{1}{3}}{1}$

Equation of line passing through

(- 1, - 1, 2)

$(-1, -1, 2)$ parallel to

L_{1}

$L_1$

\frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}

$\dfrac{x + 1}{3} = \dfrac{y + 1}{2} = \dfrac{z - 2}{1}$

vector equation is

(-^i -^j + 2^k) + λ (3^i + 2^j +^k)

$(- \hat{i} - \hat{j} + 2 \hat{k}) + \lambda ( 3 \hat{i} + 2 \hat{j} + \hat{k})$

¯ a, ¯ b

$\bar {a}, \bar {b}$ and

¯ c

$\bar {c}$ are non-zero non collinear vectors and

θ (\neq 0, π)

$\theta(\neq 0 , \pi)$ is the angle between

¯ b

$\bar {b}$ and

¯ c

$\bar {c}$ if

(¯ a \times ¯ b) \times ¯ c = \frac{1}{2} | ¯ b | ¯ c | ¯ a

$(\bar {a}\times \bar {b}) \times \bar {c}=\dfrac {1}{2} |\bar {b}|\bar {c}|\bar {a}$ . then

sin θ =

$\sin \theta =$

0%
$\sqrt{\dfrac{2}{3}}$
0%
$\dfrac{\sqrt{3}}{2}$
0%
$\dfrac{4\sqrt{2}}{3}$
0%
$\dfrac{2\sqrt{2}}{3}$

Explanation

We have

(\to a \times \to b) \times \to c = \frac{1}{2} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a

$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a$

\to c \times (\to a \times \to b) = \frac{1}{2} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a

$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a$

- [(\to c . \to b) \to a - (\to c . \to a) \to b] = \frac{1}{2} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a

$- \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right] = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a$

(\to c . \to a) \to b - (\to c . \to b) \to a = \frac{1}{2} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a

$\left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a$

\to c . \to a = 0

$\overrightarrow c .\overrightarrow a = 0$

\to c . \to a = \frac{- 1}{2} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣

$\overrightarrow c .\overrightarrow a = \frac{{ - 1}}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|$

cos θ = \frac{- 1}{2}

$\cos \theta = \frac{{ - 1}}{2}$

\Rightarrow θ = \frac{2 π}{3}

$\Rightarrow \theta = \frac{{2\pi }}{3}$

∴ sin θ = \frac{\sqrt{3}}{2}

$\therefore \sin \theta = \frac{{\sqrt 3 }}{2}$

Hence,

B

$B$ is the correct answer.

A = (1, 2, - 1), B = (2, 0, 3), C = (3, - 1, 2)

$A=(1,2,-1), B=(2,0,3), C=(3,-1,2)$ then the angle between

¯ ¯¯¯¯¯¯ ¯ A B

$\overline { AB }$ and

¯ ¯¯¯¯¯¯ ¯ A C

$\overline { AC }$ is

0%
${0}^{o}$
0%
${90}^{o}$
0%
$\cos ^{ -1 }{ \left( \cfrac { 20 }{ \sqrt { 21 } \sqrt { 22 } } \right) }$
0%
$\cos ^{ -1 }{ \left( \cfrac { 15 }{ \sqrt { 21 } \sqrt { 11 } } \right) }$

Explanation

Given

A (1, 2, - 1), B = (2, 0, 3), C = (3, - 1, 2)

$A(1,2,-1),B=(2,0,3),C=(3,-1,2)$

Drs of

A B = (1, - 2, 4)

$AB=(1,-2,4)$

Drs of

A C = (2, - 3, 3)

$AC=(2,-3,3)$

cos θ = \frac{2 + 6 + 12}{\sqrt{1 + 4 + 16} \sqrt{4 + 9 + 9}}

$\cos \theta =\cfrac{2+6+12}{\sqrt{1+4+16}\sqrt{4+9+9}}$

\Rightarrow

$\Rightarrow$

cos θ = \frac{20}{\sqrt{21} \sqrt{22}}

$\cos \theta =\cfrac{20}{\sqrt {21}\sqrt {22}}$

\Rightarrow

$\Rightarrow$

θ = {cos}^{- 1} (\frac{20}{\sqrt{21} \sqrt{22}})

$\theta =\cos ^{ -1 }{ \left( \cfrac { 20 }{ \sqrt { 21 } \sqrt { 22 } } \right) }$

A line d.c's proportional to

(2, 1, 2)

$(2,1,2)$ meets each of the lines

x = y + a = z

$x=y+a=z$ and

x + a = 2 y = 2 z

$x+a=2y=2z$ . Then the coordinates of each of the points of intersection are given by

0%
$(3a,2a,3a); (a,a,2a)$
0%
$(3a,2a,3a); (a,a,a)$
0%
$(3a,3a,3a); (a,a,a)$
0%
$(2a,3a,3a); (2a,a,a)$

Explanation

L_{1} : \frac{x}{1} = \frac{y + a}{1} = \frac{z}{1}

$L_1 : \dfrac{x}{1} = \dfrac{y + a}{1} = \dfrac{z}{1}$

L_{2} : \frac{x + a}{2} = \frac{y}{1} = \frac{z}{1}

$L_2 : \dfrac{x + a}{2} = \dfrac{y}{1} = \dfrac{z}{1}$

Any point on

L_{1} A (k_{1}, k_{1} - 1, k_{1})

$L_1 \, A (k_1, k_1 - 1, k_1)$

Any point on

L_{2} B (2 k_{2} - 1, k_{2}, k_{2})

$L_2 \, B (2k_2 - 1, k_2, k_2)$

Drs of

A B = (k_{1} - 2 k_{2} + a, k_{1} - k_{2} - a, k_{1} - k - 2)

$AB = (k_1 - 2k_2 + a , k_1 - k_2 - a , k_1 - k-2)$

= \frac{k_{1} - 2 k_{2} + a}{2} = \frac{k_{1} - k_{2} - a}{1} = \frac{k_{1} - k_{2}}{2}

$= \dfrac{k_1 - 2k_2 + a}{2} = \dfrac{k_1 - k_2 - a}{1} = \dfrac{k_1 - k_2}{2}$

\Rightarrow k_{1} - 2 k_{2} + a = 2 k_{1} - 2 k_{2} - 2 a

$\Rightarrow k_1 - 2k_2 + a = 2k_1 - 2k_2 - 2a$

\Rightarrow 3 a = k_{1}

$\Rightarrow 3a = k_1$

∴ 2 k_{1} - 2 k_{2} - 2 a = k_{1} - k_{2}

$\therefore 2k_1 - 2k_2 - 2a = k_1 - k_2$

\Rightarrow 2 k_{1} - k_{1} - k_{2} = 2 a

$\Rightarrow 2k_1 - k_1 - k_2 = 2a$

\Rightarrow k_{1} - k_{2} = 2 a

$\Rightarrow k_1 - k_2 = 2a$

\Rightarrow k_{2} = a

$\Rightarrow k_2 = a$

∴ A (3 a, 2 a, 3 a)

$\therefore A (3a , 2a , 3a)$

B (a, a, a)

$B (a, a, a)$

1238412_1195289_ans_22ccc1e13bd0471ea44a7472ff4bf702.JPG

\frac{x - 14}{l} = \frac{y - 2}{m} = \frac{z + 1}{n}

$\frac{x-14}{l}=\frac{y-2}{m}=\frac{z+1}{n}$ is the equation of the line through (1,2,-1) and (-1,0,1), then (l,m,n) is

0%
(-1,0,1)
0%
(1,1,-1)
0%
(1,2,-1)
0%
(0,1,0)

A = (1, 2, 3), B = (2, 10, 1), Q

$A = (1,2,3) , B = (2,10,1), Q$ are collinear points and

Q_{x} = - 1

$Q_{x}=-1$ then

Q_{z}

$Q_{z}$ is

0%
$-3$
0%
$7$
0%
$-14$
0%
$-7$

The direction cosines to two lines at right angles are (1,2,3) and (-2,

\frac{1}{2}

$\frac{1}{2}$ ,

\frac{1}{3}

$\frac{1}{3}$ ), then the direction cosine perpendicular to both given lines are:

0%
$\sqrt{\frac{25}{2198}}$ , $\sqrt{\frac{19}{2198}}$ , $\sqrt{\frac{729}{2198}}$
0%
$\sqrt{\frac{24}{2198}}$ , $\sqrt{\frac{38}{2198}}$ , $\sqrt{\frac{730}{2198}}$
0%
$\frac{1}{3}$ ,-2, $\frac{-7}{2}$
0%
None of the above

Explanation

Direction cosines of

L_{1} = (1, 2, 3)

${ L }_{ 1 }=\left( 1,2,3 \right)$

Direction cosines of

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

${ L }_{ 2 }=\left( -2,\cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 3 } \right)$

Let direction ratios of required line be a,b,c then

a + 2 b + 3 c = 0 \to 1

$a+2b+3c=0\rightarrow 1$

and

- 2 a + \frac{b}{2} + \frac{c}{3} = 0

$-2a+\cfrac { b }{ 2 } +\cfrac { c }{ 3 } =0$

\Rightarrow - 12 a + 3 b + 2 c = 0 \to 2

$\Rightarrow -12a+3b+2c=0\rightarrow 2$

Solving

1

$1$ and

2

$2$

a + 2 b + 3 c = 0

$a+2b+3c=0$

- 12 a + 3 b + 2 c = 0

$-12a+3b+2c=0$

\Rightarrow \frac{a}{4 - 9} = \frac{b}{- 36 - 2} = \frac{c}{3 + 24}

$\Rightarrow \cfrac { a }{ 4-9 } =\cfrac { b }{ -36-2 } =\cfrac { c }{ 3+24 }$

\Rightarrow \frac{a}{+ 5} = \frac{b}{+ 38} = \frac{c}{- 27}

$\Rightarrow \cfrac { a }{ +5 } =\cfrac { b }{ +38 } =\cfrac { c }{ -27 }$

\Rightarrow a = + 5, b = + 38, c = - 27

$\Rightarrow a=+5,b=+38,c=-27$

∴

$\therefore$ Direction cosines are

\frac{5}{\sqrt{5^{2} + 38^{2} + {(27)}^{2}}}, \frac{38}{\sqrt{5^{2} + 38^{2} + {(- 27)}^{2}}}, \frac{- 27}{\sqrt{5^{2} + 38^{2} + {(- 27)}^{2}}}

$\cfrac { 5 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( 27 \right) }^{ 2 } } } ,\cfrac { 38 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( -27 \right) }^{ 2 } } } ,\cfrac { -27 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( -27 \right) }^{ 2 } } }$

$\cfrac { 5 }{ \sqrt { 2198 } } ,\cfrac { 38 }{ \sqrt { 2198 } } ,\cfrac { -27 }{ \sqrt { 2198 } }$

The direction cosines of a vector

$\overrightarrow { A }$ are

$\cos \alpha = \frac {4} { 5 \sqrt {2}} , \cos\beta =\frac { 1 }{ \sqrt { 2 } } , \cos\gamma =\frac { 3 }{ 5\sqrt { 2 } }$ then, the vector

$\overrightarrow {A}$ is

0%
$4\hat { i } +\hat { j } +3\hat { k }$
0%
$4\hat { i } +5\hat { j } +3\hat { k }$
0%
$4\hat { i } -5\hat { j } +3\hat { k }$
0%
$4\hat { i } -\hat { j } -3\hat { k }$

$\bar { a }, \bar { b }, \bar { c }$ are non-coplaner vector , then the vectors

$2\bar { a }- 4\bar { b }+ 4\bar { c }, \bar { a }- 2\bar { b }+ 4\bar { c }$ and

$-\bar { a }+ 2\bar { b }+ 4\bar { c }$ are parellel.

0%
True
0%
False

The angle between the lines

$\frac{{x - 2}}{3} = \frac{{y + 1}}{{ - 2}},z = 2$ and

$\frac{{x - 1}}{1} = \frac{{2y + 3}}{3} = \frac{{z + 5}}{2}$ is equal to

0%
$\pi /2$
0%
$\pi /3$
0%
$\pi /6$
0%
none of these

The direction ratios of the line joining the points

$(4, 3, -5)$ and

$(-2, 1, -8)$ are

0%
$\dfrac{6}{7}, \dfrac{2}{7}, \dfrac{3}{7}$
0%
$6, 2, 3$
0%
$5,8,0$
0%
$3,7,9$

Explanation

Given points

$P(4,3,-5)$

$Q(-2,1,-8)$

Direction ratios of

$PQ$

$=$ Position vector of

$P-$ Position vector of

$Q$

$=4-(-2),3-1,-5-(-8)$

$=6,2,3$

$\left(\dfrac {1}{2},\dfrac {1}{3},n\right)$ are the direction cosines of a line then the value of

$n$ is

0%
$\dfrac {\sqrt {23}}{6}$
0%
$\dfrac {23}{6}$
0%
$\dfrac {2}{3}$
0%
$\dfrac {3}{2}$

Explanation

Given direction cosines are

$\cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 3 } ,n$

we have that if

$l,m,n$ are direction cosines of a line then

${ l }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1$

$\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 9 } +{ n }^{ 2 }=1\\ \Rightarrow { n }^{ 2 }=1-(\cfrac { 13 }{ 36 } )=\cfrac { 23 }{ 36 } \\ n=\cfrac { \sqrt { 23 } }{ 6 }$

The angle between the pair of lines with direction ratios (1, 1, 2) and

$(\sqrt{3} - 1, -\sqrt{3} - 1, 4)$ is

0%
$30^o$
0%
$45^o$
0%
$60^o$
0%
$90^o$

Explanation

$cos \theta = \dfrac{1(\sqrt{3} - 1) + 1 (-\sqrt{3} - 1) + 2(4)}{\sqrt{1^2 + 1^2 + 2^2} \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2}}$

$= \dfrac{\sqrt{3} - 1 - \sqrt{3} - 1 + 8}{\sqrt{6}. \sqrt{4 - 2 \sqrt{3} + 4 + 2\sqrt{3} + 16}}$

$= \dfrac{6}{\sqrt{6} . \sqrt{24}}$

$= \dfrac{6}{12}$

$= \dfrac{1}{2}$

$\theta = 60^o$

A line makes angles

$\alpha,\beta,\gamma,\delta$ with the four diagonals of a cube then

$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta$ is equal to

0%
$1$
0%
$4/3$
0%
$3/4$
0%
$4/5$

Explanation

REF.Image

Take 'O' as a corner

$OA,OB,OC$ are 3 edges through the axes

Let

$OA=OB=OC=a$

coordinates of

$O=(o,o,o)$

$A(a,o,o) B(o,a,o) C(o,o,a)$

$P(a,a,o) L(o,a,a) M(a,o,a) N(a,a,o)$

The four diagonals

$OP, AL, BM, CN$

Direction cosine of

$OP : a-o,a-o,a-o = a,a,a = 1,1,1$

Direction cosine of

$AL : o-a,a-o,a-o = -a,a,a = -1,1,1$

Direction cosine of

$BM : a-o,o-a,a-o = a,-a,a = 1,-1,1$

Direction cosine of

$CN : a-o,a-o,o-a = a,a,-a = 1,1,-1$

$\therefore$ DC's of OP are

$\displaystyle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

DC's of AL are

$\displaystyle\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

DC's of BM are

$\displaystyle\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

DC's of CN are

$\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}}$

Let l,m,n be dc's of line and line makes angle

$\alpha$

with OP :-

$\displaystyle \cos \alpha = l(\frac{1}{\sqrt{3}})+m(\frac{1}{\sqrt{3}})+n(\frac{1}{\sqrt{3}})=\frac{l+m+n}{\sqrt{3}}$

Similarly

$\cos\, \beta =\dfrac{-l+m+n}{\sqrt{3}}$

$\cos \delta = \dfrac{l+m-n}{\sqrt{3}}$

$\cos\gamma =\dfrac{l-m+n}{\sqrt{3}}$

suaring and adding all the four

i.e ;

$\cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta$

$= \dfrac{1}{3}[(l+m+n)^{2}+(-l+m+n)^{2}+(l-m+n)^{2}+(l+m-n)^{2}]$

$=\dfrac{1}{3}[4l^{2}+4m^{2}+4n^{2}]=\dfrac{4}{3}(l^{2}+m^{2}+n^{2})$

$[\because l^{2}+m^{2}+n^{2}=1]=\dfrac{4}{3}$

$\therefore \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta =\dfrac{4}{3}$

1369886_1229776_ans_0159f740952e448f9cbe7f71fecbb828.JPG

The direction ratios of the line, given by the planes x - y + z - 5 = 0, x - 3y - 6 = 0 are

0%
(3, 1, -2)
0%
(2, -4, 1)
0%
(1,-1, 1)
0%
(0,2,1)

Explanation

Given,

$x – y + z – 5 = 0 = x – 3y – 6$

$\Rightarrow x – y + z – 5 = 0$

$x – 3y – 6 = 0$

$\Rightarrow x – y + z – 5 = 0$ (1)

$x = 3y + 6$ (2)

From (1) and (2) we get,

$3y + 6 – y + z – 5 = 0$

$2y + z + 1 = 0$

$y =\dfrac{-z-1}{2}$

Also,

$y=\dfrac{x-6}{3}$ From (2)

$\therefore \dfrac{x-6}{3}=y=\dfrac{-z-1}{2}$

So, the given equation can be re-written as

$\dfrac{x-6}{3}=\dfrac{y}{1}=\dfrac{z+1}{-2}$

Hence the direction ratios of the given line are proportional to

$3,1,-2$

$\overline { O A } = 3 \overline { i } + \overline { j } - \overline { k }$ ,

$| \overline { A B } | = 2 \sqrt { 6 }$ and AB has the direction ratios 1, -1 , 2 then

$| O B | =$

0%
$\sqrt { 35 }$
0%
$\sqrt { 41 }$
0%
$\sqrt { 26 }$
0%
$\sqrt { 55 }$

Explanation

$\vec{OA}=3\hat{i}+\hat{j}-\hat{k}$

$|\vec{AB}|=2\sqrt{6}$ has dr's

$(1, -1, 2)$

Unit vector with above dr's are

$\vec{AB}=\left(\dfrac{1}{\sqrt{6}}\hat{i}-\dfrac{1}{\sqrt{6}}\hat{j}+\dfrac{2}{\sqrt{6}}\hat{k}\right)(2\sqrt{6})$

$\vec{AB}=2\hat{i}-2\hat{j}+4\hat{k}$

$\vec{AB}=\vec{OB}-\vec{OA}$

$\vec{OB}=(2\hat{i}-2\hat{j}+4\hat{k})+(3\hat{i}+\hat{j}-\hat{k})$

$\vec{OB}=(5\hat{i}-\hat{j}+3\hat{k})$

$|OB|=\sqrt{25+1+9}=\sqrt{35}$ .

The direction cosines of a vector A are

$\cos { \alpha } =\frac { 4 }{ 5\sqrt { 2 } } ,$

$cos \beta =\frac { 1 }{ \sqrt { 2 } } ,$ and

$cos \gamma = \frac{ 3 }{ 5\sqrt { 2 } } ,$ then vector A is

0%
$4i+j+3k$
0%
$4i+5j+3k$
0%
4i-5j-3k
0%
none

Explanation

$\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}$

$\cos \alpha = \dfrac{a}{\sqrt{a^2+ b^2+c^2}}$

$\cos \beta = \dfrac{b}{\sqrt{a^2 + b^2 + c^2}}$

$\cos \gamma = \dfrac{c}{\sqrt{a^2 + b^2 + c^2}}$

$\sqrt{4^2 + 5^2 + 3^2} = \sqrt{16 +25+ 9} = \sqrt{50} = 5\sqrt{2}$

$\vec{A} = 4\hat{i} + 5\hat{j} + 3\hat{k}$

The vector

$a = \alpha 1 + 2 j + \beta k$ lies in the plane of the vectors

$b = i + jt$ and

$c = j + k$ and bisects the angle between

$b$ and

$c$ . Then which one of the following gives possible values

$\alpha$ and

$\beta$ .

0%
$\alpha = 1 , \beta = 2$
0%
$\alpha = 2 , \beta = 1$
0%
$\alpha = 1 , \beta = 1$
0%
$\alpha = 2 , \beta = 2$

Let

$l_{1},\ m_{1},\ n_{1};\ l_{2},\ m_{2},\ n_{2};\ l_{3},\ m_{3},\ n_{3}$ be the direction cosines of three mutually perpendicular line then

$\begin{vmatrix} { l }_{ 1 } & m_{ 1 } & n_{ 1 } \\ { l }_{ 2 } & m_{ 2 } & n_{ 2 } \\ { l }_{ 3 } & m_{ 3 } & n_{ 3 } \end{vmatrix}$

0%
$0$
0%
$\pm 1$
0%
$\pm 2$
0%
$\pm \dfrac{1}{2}$

The point where

$\vec{ x }$ which is perpendicular to

$(2,-3,1)$ and

$(1,-2,3)$ and which satisfies the condition

$\vec { x } \cdot ( \hat { i } + 2 \hat { j } - 7 \hat{ k } ) = 10$

0%
$\left(3,5,1\right)$
0%
$\left(7,-5,1\right)$
0%
$\left(3,-5,1\right)$
0%
$\left(7,5,1\right)$

Explanation

$\bar{x} = (a , b, c)$

$\bar{x} \bot (2, -3, 1) \, \& \, x \bot (1, -2 , 3)$

$\Rightarrow x .(2, -3, 1) = 0$ &

$x. (1, -2, 3) = 0$

$\Rightarrow 2a - 3b + c = 0$ __(1) &

$a - 2b + 3c = 0$ __(2)

$\bar{x} . (\hat{i} + 2 \hat{j} - 7 \hat{k} ) = 10$

$\Rightarrow a + 2b - 7c = 10$ __(3)

$\left.\begin{matrix}(1) \Rightarrow c = 3b - 2a\\(2) \Rightarrow a = 2b - 3c \end{matrix}\right\}$ Put in (3)

$a + 2b - 7c = 10$

$2b - 3c + 2b - 7c = 10$

$\Rightarrow 4b - 10 c = 10$ __(4)

(1)

$\Rightarrow a = \dfrac{-c + 3b}{2}$

$a = \dfrac{3b - c}{2} = 2b - 3c$

$\Rightarrow 3b - c = 4b - 6c$

$\Rightarrow b = 5c$ __(5)

from (4) & (5)

$c = 1 , b = 5, a = 7$

$\Rightarrow \bar{x} = (7, 5 , 1)$

The equation of the plane through

$\left(0,-5,1\right)$ which is perpendicular to the planes

$2x+4y+2z+3=0$ ,

$2x+5y+3z+4=0$ is

0%
$x+y+z=6$
0%
$x-y+z=6$
0%
$x-y-z=6$
0%
$x+y+z+6=0$

$A(p,q,r)$ and

$B=(p\prime ,q\prime ,r\prime )$ are two points on the line

$\lambda x=\mu y=yz$ such that

$OA=3,OB=4$ then

$pp\prime +qq\prime +rr\prime$ is equal to

0%
$7$
0%
$12$
0%
$5$
0%
$None$ $of$ $these$

The angle between the lines whose de's satisfy the equation

$l+m+m=0$ and

$l^2+m^{2}-n^{2}=0$ is

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{4}$

Explanation

Given that the equations

$l+m+n=0$ ………..

$(1)$

$l+m=-n$

$\Rightarrow -(l+m)=n$

and

$l^2+m^2+n^2=0$ ……….

$(2)$

Put the value of n in equation

$(2)$

$l^2+m^2+n^2=0$

$\Rightarrow l^2+m^2-(-(l+m))^2=0$

$\Rightarrow l^2+m^2-(l^2+m^2-2ml)=0$

$\Rightarrow l^2+m^2-l^2-m^2+2ml=0$

$\Rightarrow 2ml0$

$\Rightarrow ml=0$

$\Rightarrow m=0, l=0$

Let us put

$m=0$ in equation

$(3)$

$l+o+n=0$

$l=-n$

Hence, direction rates

$(l, m, o)=(1, 0, -1)$

Let us put

$l=0$ , we get

$m=-n$

Here, direction ratios

$(l, m, n)=(0, 1, -1)$

we know that,

$\cos\theta =\dfrac{\vec{b_1}\cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}|}$

$=\dfrac{(1, 0, -1)\cdot (0, 1, -1)}{\sqrt{1^2+0^2+(-1)^2}\sqrt{0^2+1^2+(-1)^2}}$

$=\dfrac{1}{\sqrt{2}\sqrt{2}}$

$\cos\theta =\dfrac{1}{2}$

$\cos\theta =\cos\dfrac{\pi}{3}$

$\theta =\dfrac{\pi}{3}$

Hence, this is the answer.

1197487_1242755_ans_31da2c2be3324d8bb45d4577225149df.png

The angle between the lines, whose direction ratios are

$1,1,2$ and

$\sqrt { 3 } - 1 , - \sqrt { 3 } - 1,4 ,$ is

0%
${45} ^ { \circ }$
0%
${30} ^ { \circ }$
0%
${60} ^ { \circ }$
0%
${90} ^ { \circ }$

Explanation

The direction ratios are

$(1, 1, 2)$ and

$(\sqrt{3}-1, -\sqrt{3}-1, 4)$

Angle between them is given by

$\cos\theta =\dfrac{1(\sqrt{3}-1)+1(-\sqrt{3}-1)+2/4)}{\sqrt{1^2+1^2+2^2}\sqrt{(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2+4^2}}$

$=\dfrac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6}\sqrt{8+16}}$

$=\dfrac{6}{\sqrt{6}\sqrt{24}}=\dfrac{6}{\sqrt{144}}$

$=\dfrac{6}{12}$

$=\dfrac{1}{2}$

$\cos\theta =\dfrac{1}{2}$

$\Rightarrow \theta =\dfrac{\pi}{3}=60^o$ .

The directions cosines of the line which is perpedicular to the lines whose direction cosines are proportional to (1, -1, 2) and (2,-1,-1) are:

0%
$\dfrac { 1 }{ \sqrt { 35 } } ,-\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } }$
0%
$-\dfrac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } \dfrac { 3 }{ \sqrt { 35 } }$
0%
$\dfrac { 1 }{ \sqrt { 35 } } ,\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } }$
0%
None of these

If a plane passes through the point

$(1, 1, 1)$ and is perpendicular to the line

$\dfrac{x-1}{3}=\dfrac{y-1}{0}=\dfrac{z-1}{4}$ then its perpendicular distance from the origin is

0%
$\dfrac{3}{4}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{7}{5}$
0%
$1$

Explanation

$\begin{array}{l} Let\, eq{ u^{ n } }\, of\, plane\, be\, :\, a\left( { x-1 } \right) +b\left( { y-1 } \right) +\left( { z-1 } \right) =0 \\ 3\left( { x-1 } \right) +0\left( { y-1 } \right) +4\left( { z-1 } \right) =0 \\ 3x+4z=7 \\ perpendicular\, \, dis\tan ce\, from\, \, origin\, \, d=\frac { 7 }{ 5 } \\ Hence,\, the\, option\, C\, is\, correct\, answer. \end{array}$

The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as

$l_{1},m_{1},n_{1};l_{2},m_{2},n_{2}$ and

$l_{3},m_{3},n_{3}$ are

0%
$l_{1}+ l_{2}+ l_{3},m_{1}+m_{2}+m_{3},n_{1}+n_{2}+n_{3}$
0%
$\dfrac{l_{1}+l_{2}+l_{3}}{\sqrt{3}},\dfrac{m_{1}+m_{2}+m_{3}}{\sqrt{3}},\dfrac{n_{1}+n_{2}+n_{3}}{\sqrt{3}}$
0%
$\dfrac{l_{1}+l_{2}+l_{3}}{3},\dfrac{m_{1}+m_{2}+m_{3}}{3},\dfrac{n_{1}+n_{2}+n_{3}}{3}$
0%
$None\ of\ these$

$l=m=n=1$ represent the direction cosines of the

0%
$x-$ axis
0%
$y-$ axis
0%
$z-$ axis
0%
$none\ of\ these$

Explanation

$l=m=n=1$

$(1,1,1)$ represent direction ratio of line equally inclined to

$x,y\, \&\,zaxis$

Hence,

option

$D$ is correct answer.

If the points (p. 0), (0, q) and (1, 1) are collinear then

$\dfrac { 1 }{ p } +\dfrac { 1 }{ q }$ is equal to

0%
-1
0%
1
0%
2
0%
0

Explanation

If the area of triangle is zero, then the points are collinear.

Points are collinear.

Point are

$(p, o)(o, q)(i,q)$

$\Delta =\dfrac {1}{2}[p(q-1)-0(1-0)+1(0-q)]$

$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$

$\Rightarrow \dfrac{1}{2}[pq-(p+q)]=o$

$\Rightarrow pq=p+q\Rightarrow \dfrac {p+q}{pq}=1$

$\Rightarrow \dfrac{1}{p}+\dfrac{1}{q}=1$

The direction ratios of the line

$x-y+z-5=\quad 0\quad =\quad x-3y-6\quad are$

0%
$3,1,-2$
0%
$2,-4,1$
0%
$\frac { 3 }{ \sqrt { 14 } } ,\frac { 1 }{ \sqrt { 14 } } ,\frac { -2 }{ \sqrt { 14 } }$
0%
$\frac { 2 }{ \sqrt { 41 } } ,\frac { -4 }{ \sqrt { 41 } } ,\frac { 1 }{ \sqrt { 41 } }$

Explanation

$x-y+z-5=0=x-3y-6$

$x-y+z-5=0$ --(I)

$x-3y-6=0$ --(II)

The Direction ratios of the line is

$(\hat{i} - \hat{j} + \hat{k})\times(\hat{i} -3\hat{j})$

$3\hat{i}+\hat{j}-2\hat{k}$

Hence,the direction ratios is (3,1,-2).

Hence, Option A is correct

The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as

$l_1$ ,

$m_1$ ,

$n_1$ :

$l_2$ ,

$m_2$ ,

$n_2$ and

$l_3$ ,

$m_3$ ,

$n_3$ are

0%
$l_1$ + $l_2$ + $l_3$ , $m_1$ + $m_2$ + $m_3$ , $n_1$ + $n_2$ + $n_3$
0%
$\dfrac{l_1+l_2+l_3}{\sqrt{3}}$ , $\dfrac{m_1+m_2+m_3}{\sqrt{3}}$ , $\dfrac{n_1+n_2+n_3}{\sqrt{3}}$
0%
$\dfrac{l_1+l_2+l_3}{3}$ , $\dfrac{m_1+m_2+m_3}{3}$ , $\dfrac{n_1+n_2+n_3}{3}$
0%
None of these

$A(3\hat { i } +2\hat { j } +3\hat { k } ),B(-\hat { i } -\hat { j } +8\hat { k } ),C(-4\hat { i } +4\hat { j } +6\hat { k } )$ are the vertices of a triangle then the equation of the line passing through the circumcentre and parallel to

$\vec { A B }$ is

0%
$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$
0%
$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$
0%
$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } -\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$
0%
$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } -\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$

The cartesian from of equation a line passing through the point position vector

$2\hat{i}-\hat{j}+2\hat{k}$ and is in the direction of

$-2\hat{i}+\hat{j}+\hat{k}$ , is

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$\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}$
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$\dfrac{x+4}{-2}=\dfrac{y-1}{1}=\dfrac{z+2}{1}$
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$\dfrac{x+2}{4}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}$
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$None \ of \ these$

Explanation

Equation of a line passing through a point with position vector

$\vec{a}$ and parallel to

$\vec{b}$ is

$\vec{r}=\vec{a}+\lambda\vec{b}$

Here

$\vec{a}= 2\hat{i}−\hat{j}+2\hat{k}$ and

$\vec{b}=−2\hat{i}+\hat{j}+\hat{k}$

So,

$\vec{r}=2\hat{i}−\hat{j}+2\hat{k}+\lambda\left(−2\hat{i}+\hat{j}+\hat{k}\right)$

Equation of the line in vector form is

$2\hat{i}−\hat{j}+2\hat{k}+\lambda\left(−2\hat{i}+\hat{j}+\hat{k}\right)$

Since the line passes through a point with position vector

$2\hat{i}−\hat{j}+2\hat{k}$

$\therefore {x}_{1}=2,\,{y}_{1}=-1,\,{z}_{1}=2$

Also, line is in the direction of

$−2\hat{i}+\hat{j}+\hat{k}$

Direction ratios :

$a=-2,\,b=1,\,c=1$

Equation of line in cartesian form is

$\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}$

$\cos { \alpha ,\quad \cos { \beta ,\quad \cos { \gamma } } }$ are the direction cosine of a line, then find the value of

${ cos }^{ 2 }\alpha +\left( \cos { \beta +\sin { \gamma } } \right)$

$\left( \cos { \beta - \sin { \gamma } } \right)$

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$2$
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$0$
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$-1$
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$1$

$\dfrac { x - 2 } { 1 } = \dfrac { y - 3 } { 1 } = \dfrac { z - 4 } { - 1 }$ &

$\dfrac { x - 1 } { k } = \dfrac { y - 4 } { 2 } = \dfrac { z - 5 } { 2 }$ are coplanar then k=?

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any value
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exactly one value
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exactly $2$ values
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exactly $3$ values

Explanation

If this two lines are co-planar

$\begin{vmatrix} 1& 1&-1 \\ k & 2 & 2\\ 2-1&3-4&4-5\end{vmatrix}=0$

$\begin{vmatrix} 1&1&1 \\k&2 &2\\ 1&-1&-1\end{vmatrix}=0$

$1(-2+2)-1(-k-2)-1(-k-2)=0$

$k + 2+ k + 2 = 0$

$k = -2$

The plane through (1, 1, 1) (1, -1, 1) and (-7, -3, -5) is

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Parallel to x-axis.
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Parallel to y-axis.
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Perpendicular to y-axis.
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Perpendicular to x-axis.

Direction ratio of line given by

$\dfrac { x-1 }{ 3 } =\dfrac { 6-2y }{ 10 } =\dfrac { 1-z }{ -7 }$ are:

0%
$<3,10,-7>$
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$<3,-5,7>$
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$<3,5,7>$
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$<3,5,-7>$

Explanation

$\cfrac{x-1}{3} = \cfrac{6 - 2y}{10} = \cfrac{1 - z}{-7}$

$\cfrac{x-1}{3} = \cfrac{2y - 6}{-10} = \cfrac{z - 1}{- \left( -7 \right)}$

$\cfrac{x - 1}{3} = \cfrac{y - 3}{\left( \cfrac{-10}{2} \right)} = \cfrac{z - 1}{7}$

$\cfrac{x-1}{3} = \cfrac{y - 3}{-5} = \cfrac{z-1}{7}$

Therefore,

Direction ratios are

$3, -5, 7$

A normal to the plane

$x=2$ is...

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$(0,1,1)$
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$(2,0,2)$
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$(1,0,0)$
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$(0,1,0)$

Explanation

The plane

$x=2$ is perpendicular to

$x axis$

So the angle is

$\dfrac{\pi}{2},\cos \dfrac{\pi}{2}=0$

The plane

$x=2$ is parallel to both

$y axis$ and

$z axis$

So the angle is

$0,\cos 0=1$

So the Drs are

$(0,1,1)$

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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