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CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Three Dimensional Geometry
Quiz 8
ABC is a triangle where $$A = ( 2,3,5 ) , B = ( - 1,2,2 )$$ and $$C (\lambda,5 , \mu )$$ if the median through A is equally inclined to the positive axis then $$\lambda + \mu$$ is
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7
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6
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15
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9
Explanation
$$AD$$ is median $$D\left( {\frac{{\lambda - 1}}{2},\frac{7}{2},\frac{{\mu + 2}}{2}} \right)$$
$$\overrightarrow {AD} = \left( {\frac{{\lambda - 1}}{2} - 2,\frac{7}{2} - 3,\frac{{\mu + 2}}{2} - 5} \right)$$
$$\overrightarrow {AD} $$ is equally in lined. $$ \therefore$$paralle to $$(1,1,1)$$
$$\begin{array}{l} \frac { { \lambda -5 } }{ 2 } =\frac { 1 }{ 2 } =\frac { { \mu +8 } }{ 2 } \\ \lambda =6 \\ \mu =9 \\ \lambda +\mu =15 \end{array}$$
Then,
Option $$C$$ is correct answer.
Projection of a vector on $$3$$ coordinate axes are $$6, - 3, 2$$ respectively. Then DC's of vector are-
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$$6, - 3, 2$$
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$$\dfrac{6}{5},\dfrac{{ - 3}}{5},\dfrac{2}{5}$$
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$$\dfrac{-6}{7},\dfrac{{ - 3}}{2},\dfrac{2}{7}$$
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$$\dfrac{ 6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$$
Explanation
Projection of a vector on coordinate axis are $$X_2-X_1, Y_2-Y_1, Z_2-Z_1$$
$$X_2-X_1=6, Y_2-Y_1=-3, Z_2-Z_1=2$$
$$\sqrt {{{\left( {{X_2} - {X_1}} \right)}^2} + {{\left( {{Y_2} - {Y_1}} \right)}^2} + {{\left( {{Z_2} - {Z_1}} \right)}^2}} = \sqrt {36 + 9 + 4} = 7$$
The Direction cosines of the vector are $$\dfrac{6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$$
If the foot of the perpendicular from $$(0,0,0)$$ to a plane is $$P(1,2,2)$$. Then, the equation of the plane is
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$$-x+2y+8z-9=0$$
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$$x+2y+2z-9=0$$
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$$x+y+z-5=0$$
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$$x+2y-3z+3=0$$
Explanation
Equation of normal $$\Rightarrow \ \hat i+2\hat j+2\hat k$$
equation of plane $$\Rightarrow $$
$$(x-1)1+(xy-2)2+(z-2)2=0$$
$$x-1+2y-4+2z-4=0$$
$$x+2y+2z-9=0$$
$$P\left(1,1,1\right)$$ and $$Q\left(\lambda,\lambda,\lambda\right)$$ are two points in the space such that $$PQ=\sqrt{27},$$ the value of $$\lambda$$ can be
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$$-4$$
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$$-2$$
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$$2$$
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$$0$$
Explanation
$${PQ}^{2}={\left(\lambda-1\right)}^{2}+{\left(\lambda-1\right)}^{2}+{\left(\lambda-1\right)}^{2}$$
$$\Rightarrow\,3{\left(\lambda-1\right)}^{2}=27$$
$$\Rightarrow\,{\left(\lambda-1\right)}^{2}=9$$
$$\Rightarrow\,\lambda-1=\pm\,3$$
$$\Rightarrow\,\lambda=1\pm\,3$$
$$\Rightarrow\,\lambda=1+3,$$or $$\,1-3$$
$$\therefore\,\lambda=4,$$ or $$-2$$
$$A(2,3,7),B(-1,3,2)$$ and $$C(q,5,r)$$ are the vertices of $$\Delta ABC$$. If the median through A is equally inclined to the coordinate axes then the coordinates of the vertex C is
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$$(7,5,14)$$
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$$(7,5,12)$$
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$$(7,5,10)$$
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$$(7,5,16)$$
If the points $$\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$$ are collinear, then
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$$k$$ has only one real value
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$$k$$ has two real value
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$$k$$ has no real values
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$$k$$ has infinite number of real values
Explanation
As the $$3$$ point should be collinear area of triangle termed by then should be zero
considering $$2$$ direction to be $$(\bar{a}+\bar{b})-(\bar{a}-\bar{b})=2\bar{b}$$
& $$(\bar{a}+\bar{b})-(\bar{a}+k\bar{b})=(1-k)\bar{b}$$
$$(2\bar{b})\times (1-k)\bar{b}=0$$
this will be for any value of $$k$$ as cross produced of $$2$$ linear vector $$=0$$
A line $$AB$$ in three-dimensional space males angles $${45}^{o}$$ and $${120}^{o}$$ with the positive x-axis and the positive y-axis respectively. If $$AB$$ makes an acute angle $$\theta$$ with the positive z-axis, then $$\theta$$ equal
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$${45}^{o}$$
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$${60}^{o}$$
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$${75}^{o}$$
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$${30}^{o}$$
Explanation
'l', 'm', 'n' be the Direction cosine of the line
$$ l = cos \alpha = cos45^{\circ} = \dfrac{1}{\sqrt{2}}$$
$$ m = cos\beta = cos120^{\circ} = \dfrac{-1}{2}$$
$$ n = cos\theta $$
we know
$$ l^{2}+m^{2}+n^{2} = 1 $$
$$ \Rightarrow d\dfrac{1}{2}+\dfrac{1}{4}+cos^{2}\theta = 1 $$
$$ \Rightarrow cos^{2}\theta = 1-\dfrac{1}{2}- \dfrac{1}{4}$$
$$ \Rightarrow cos^{2}\theta = \dfrac{1}{4}$$
$$ \Rightarrow cos = \dfrac{1}{2}$$
$$ \therefore \theta = 60^{\circ}$$
If the points $$(\alpha, - 1), (2, 1)$$ and $$(4, 5)$$ are collinear, then find $$\alpha $$ by vector method.
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$$4$$
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$$1$$
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$$8$$
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None of these
Explanation
If there points are collinear then vectors from one to another will have scalar triple produced $$0$$.Point $$\left(\alpha,-1\right), \left(2,1\right), \left(4,5\right)$$
$$\left( 2-\alpha \right) \hat { i } +2\hat { j } -\bar { A }$$
$$2\hat { i } +4\hat { j } -\bar { B }$$
$$\left( 4-\alpha \right) \hat { i } +6\hat { j } -\bar { C }$$
$$ \bar { A } .\left( \bar { B } \times \bar { C } \right) =0$$
$$\left( \left( 2-\alpha \right) \hat { i } +2\hat { j } \right) \left( 2\hat { i } +4\hat { j } \right) \times \left( \left( 4-\alpha \right) \hat { i } +6\hat { j } \right) \\ \left( \left( 2-\alpha \right) \hat { i } +2\hat { j } \right) .\left[ 12\hat { k } -16\hat { k } +4\alpha \hat { k } \right] =0$$
$$4\alpha =4$$
$$\alpha =1$$
Also the direction vector will be proportion
$$\left( 2-\alpha,2 \right)=\lambda\left( 4-2.5-1\right)$$
$$\left( 2-\alpha,2 \right)=\lambda\left( 2,4\right)$$
$$\lambda=\dfrac{1}{2}$$ as $$2=4\lambda$$
$$2-\alpha=1$$
$$\therefore \alpha=1$$
The Cartesian equation of line $$6x - 2 = 3y + 1 = 2z - 2$$ is given by
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$$\dfrac{{3x - 1}}{3} = \dfrac{{3y + 1}}{6} = \dfrac{{z - 1}}{3}$$
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$$\dfrac{{3x + 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$$
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$$\dfrac{{3x - 1}}{3} = \dfrac{{3y - 1}}{6} = \dfrac{{z - 1}}{3}$$
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$$\dfrac{{3x - 1}}{6} = \dfrac{{3y - 1}}{3} = \dfrac{{z - 1}}{3}$$
Explanation
Solution -
$$ 6x-2 = 3y+1 = 2z-2 $$
$$ \dfrac{6x-2}{6} = \dfrac{3y+1}{6} = \dfrac{2z-2}{6} $$
$$ \dfrac{x-\dfrac{1}{3}}{1} = \dfrac{y+\dfrac{1}{3}}{2} = \dfrac{z-1}{3} $$
DR $$ (1,2,3) $$
Point passing through line $$(1/3,-1/3,1) $$
$$ \dfrac{3x-1}{3} = \dfrac{3y+1}{6} = \dfrac{z-1}{3} $$
A is correct.
The direction ratios of the joining $$A(1,\,2,\ 1)$$ and $$(2,\ 1,\ 2)$$ are
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$$3,\ 3,\ 3$$
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$$-1,\ 1,\ -1$$
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$$3,\ 1,\ 3$$
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$$\dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}}$$
Explanation
We have given points are:-
$$A\left( {1,2,1} \right)$$
$$B\left( {2,1,2} \right)$$
Now,
$$\overrightarrow {BA} = \left( {1 - 2} \right)\widehat i + \left( {2 - 1} \right)\widehat j + \left( {1 - 2} \right)\widehat k$$
$$ = - \widehat i + \widehat j - \widehat k$$
direction ratio are
$$a=-1$$
$$b=1$$
$$c=-1$$
Hence,
option $$B$$ is correct.
The direction ratios of $$AB$$ are $$- 2, 2, 1$$ . If coordinates of A are $$( 4,1,5 )$$ and $$l( A B ) = 6$$ , then coordinates of $$ B $$ ?
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$$( 0,5 , - 7 )$$
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$$( 8 , - 3,3 )$$
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$$( 0,7,5 )$$
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$$( 8,3,3 )$$
If the lines $${L}_{1}\ and {L}_{2}$$ are given by $$\bar { r } =\left( \bar { i } +2\bar { j } -\bar { k } \right) +t\left( \bar { 2i } -3\bar { j } +\bar { k } \right) \ and\bar { r } =\left( \bar { i } +\bar { j } +\bar { k } \right) +s\left( 2\bar { i } +\bar { j } -\bar { k } \right)$$, then
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$${L}_{1}\ and {L}_{2}$$ are perpendicular
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$${L}_{1}\ and {L}_{2}$$ are parallel
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$$\left ({L}_{1},{L}_{2}\right)={45}^{o}$$
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$$\left ({L}_{1},{L}_{2}\right)={60}^{o}$$
The vector equation of line passing through the point $$(-1,-1,2)$$ and parallel to the line $$2x-2=3y+1=6z-2$$
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$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (3\hat { i } +2\hat { j } +\hat { k } )$$
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$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +6\hat { k } )$$
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$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (\hat { i } +2\hat { j } +3\hat { k } )$$
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$$(-\hat { i } -\hat { j } +2\hat { k } )+\lambda (2\hat { i } +3\hat { j } +\hat { k } )$$
Explanation
Given
$$L_1 : 2x - 2 = 3y + 1 = 6z - 2$$
$$\Rightarrow 2 (x - 1) = 3 ( y + \dfrac{1}{3}) = 6(z - \dfrac{1}{3})$$
$$\Rightarrow \dfrac{x - 1}{3} = \dfrac{y + \dfrac{1}{3}}{2} = \dfrac{z - \dfrac{1}{3}}{1}$$
Equation of line passing through $$(-1, -1, 2)$$ parallel to $$L_1$$
$$\dfrac{x + 1}{3} = \dfrac{y + 1}{2} = \dfrac{z - 2}{1}$$
vector equation is
$$(- \hat{i} - \hat{j} + 2 \hat{k}) + \lambda ( 3 \hat{i} + 2 \hat{j} + \hat{k})$$
If $$\bar {a}, \bar {b}$$ and $$\bar {c}$$ are non-zero non collinear vectors and $$\theta(\neq 0 , \pi)$$ is the angle between $$\bar {b}$$ and $$\bar {c}$$ if $$(\bar {a}\times \bar {b}) \times \bar {c}=\dfrac {1}{2} |\bar {b}|\bar {c}|\bar {a}$$. then $$\sin \theta =$$
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$$\sqrt{\dfrac{2}{3}}$$
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$$\dfrac{\sqrt{3}}{2}$$
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$$\dfrac{4\sqrt{2}}{3}$$
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$$\dfrac{2\sqrt{2}}{3}$$
Explanation
We have
$$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\overrightarrow c \times \left( {\overrightarrow a \times \overrightarrow b } \right) = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$ - \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right] = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\overrightarrow c .\overrightarrow a = 0$$
$$\overrightarrow c .\overrightarrow a = \frac{{ - 1}}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|$$
$$\cos \theta = \frac{{ - 1}}{2}$$
$$ \Rightarrow \theta = \frac{{2\pi }}{3}$$
$$\therefore \sin \theta = \frac{{\sqrt 3 }}{2}$$
Hence, $$B$$is the correct answer.
If $$A=(1,2,-1), B=(2,0,3), C=(3,-1,2)$$ then the angle between $$\overline { AB } $$ and $$\overline { AC } $$ is
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$${0}^{o}$$
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$${90}^{o}$$
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$$\cos ^{ -1 }{ \left( \cfrac { 20 }{ \sqrt { 21 } \sqrt { 22 } } \right) } $$
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$$\cos ^{ -1 }{ \left( \cfrac { 15 }{ \sqrt { 21 } \sqrt { 11 } } \right) } $$
Explanation
Given $$A(1,2,-1),B=(2,0,3),C=(3,-1,2)$$
Drs of $$AB=(1,-2,4)$$
Drs of $$AC=(2,-3,3)$$
$$\cos \theta =\cfrac{2+6+12}{\sqrt{1+4+16}\sqrt{4+9+9}}$$
$$\Rightarrow$$ $$\cos \theta =\cfrac{20}{\sqrt {21}\sqrt {22}}$$
$$\Rightarrow$$ $$\theta =\cos ^{ -1 }{ \left( \cfrac { 20 }{ \sqrt { 21 } \sqrt { 22 } } \right) } $$
A line d.c's proportional to $$(2,1,2)$$ meets each of the lines $$x=y+a=z$$ and $$x+a=2y=2z$$. Then the coordinates of each of the points of intersection are given by
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$$(3a,2a,3a); (a,a,2a)$$
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$$(3a,2a,3a); (a,a,a)$$
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$$(3a,3a,3a); (a,a,a)$$
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$$(2a,3a,3a); (2a,a,a)$$
Explanation
$$L_1 : \dfrac{x}{1} = \dfrac{y + a}{1} = \dfrac{z}{1}$$
$$L_2 : \dfrac{x + a}{2} = \dfrac{y}{1} = \dfrac{z}{1}$$
Any point on $$L_1 \, A (k_1, k_1 - 1, k_1)$$
Any point on $$L_2 \, B (2k_2 - 1, k_2, k_2)$$
Drs of $$AB = (k_1 - 2k_2 + a , k_1 - k_2 - a , k_1 - k-2)$$
$$= \dfrac{k_1 - 2k_2 + a}{2} = \dfrac{k_1 - k_2 - a}{1} = \dfrac{k_1 - k_2}{2}$$
$$\Rightarrow k_1 - 2k_2 + a = 2k_1 - 2k_2 - 2a$$
$$\Rightarrow 3a = k_1$$
$$\therefore 2k_1 - 2k_2 - 2a = k_1 - k_2$$
$$\Rightarrow 2k_1 - k_1 - k_2 = 2a$$
$$\Rightarrow k_1 - k_2 = 2a$$
$$\Rightarrow k_2 = a$$
$$\therefore A (3a , 2a , 3a)$$
$$B (a, a, a)$$
If $$\frac{x-14}{l}=\frac{y-2}{m}=\frac{z+1}{n}$$ is the equation of the line through (1,2,-1) and (-1,0,1), then (l,m,n) is
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(-1,0,1)
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(1,1,-1)
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(1,2,-1)
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(0,1,0)
If $$A = (1,2,3) , B = (2,10,1), Q$$ are collinear points and $$Q_{x}=-1$$ then $$Q_{z}$$ is
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$$-3$$
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$$7$$
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$$-14$$
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$$-7$$
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:
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$$\sqrt{\frac{25}{2198}}$$,$$\sqrt{\frac{19}{2198}}$$,$$\sqrt{\frac{729}{2198}}$$
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$$\sqrt{\frac{24}{2198}}$$,$$\sqrt{\frac{38}{2198}}$$,$$\sqrt{\frac{730}{2198}}$$
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$$\frac{1}{3}$$,-2,$$\frac{-7}{2}$$
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None of the above
Explanation
Direction cosines of $${ L }_{ 1 }=\left( 1,2,3 \right) $$
Direction cosines of $${ L }_{ 2 }=\left( -2,\cfrac { 1 }{ 2 } ,\cfrac { 1 }{ 3 } \right) $$
Let direction ratios of required line be a,b,c then
$$a+2b+3c=0\rightarrow 1$$
and $$-2a+\cfrac { b }{ 2 } +\cfrac { c }{ 3 } =0$$
$$\Rightarrow -12a+3b+2c=0\rightarrow 2$$
Solving $$1$$ and $$2$$
$$a+2b+3c=0$$
$$-12a+3b+2c=0$$
$$\Rightarrow \cfrac { a }{ 4-9 } =\cfrac { b }{ -36-2 } =\cfrac { c }{ 3+24 } $$
$$\Rightarrow \cfrac { a }{ +5 } =\cfrac { b }{ +38 } =\cfrac { c }{ -27 } $$
$$\Rightarrow a=+5,b=+38,c=-27$$
$$\therefore $$Direction cosines are
$$\cfrac { 5 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( 27 \right) }^{ 2 } } } ,\cfrac { 38 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( -27 \right) }^{ 2 } } } ,\cfrac { -27 }{ \sqrt { { 5 }^{ 2 }+{ 38 }^{ 2 }+{ \left( -27 \right) }^{ 2 } } } $$
$$\cfrac { 5 }{ \sqrt { 2198 } } ,\cfrac { 38 }{ \sqrt { 2198 } } ,\cfrac { -27 }{ \sqrt { 2198 } } $$
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
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$$ 4\hat { i } +\hat { j } +3\hat { k } $$
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$$ 4\hat { i } +5\hat { j } +3\hat { k } $$
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$$ 4\hat { i } -5\hat { j } +3\hat { k } $$
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$$ 4\hat { i } -\hat { j } -3\hat { k } $$
If $$\bar { a }, \bar { b }, \bar { c }$$ 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.
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True
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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
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$$\pi /2$$
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$$\pi /3$$
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$$\pi /6$$
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none of these
The direction ratios of the line joining the points $$(4, 3, -5)$$ and $$(-2, 1, -8)$$ are
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$$\dfrac{6}{7}, \dfrac{2}{7}, \dfrac{3}{7}$$
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$$6, 2, 3$$
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$$5,8,0$$
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$$3,7,9$$
Explanation
Given points $$P(4,3,-5)$$ $$Q(-2,1,-8)$$
Direction ratios of $$PQ$$
$$=$$Position vector of $$P-$$Position vector of $$Q$$
$$=4-(-2),3-1,-5-(-8)$$
$$=6,2,3$$
If $$\left(\dfrac {1}{2},\dfrac {1}{3},n\right)$$ are the direction cosines of a line then the value of $$n$$ is
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$$\dfrac {\sqrt {23}}{6}$$
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$$\dfrac {23}{6}$$
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$$\dfrac {2}{3}$$
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$$\dfrac {3}{2}$$
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
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$$30^o$$
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$$45^o$$
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$$60^o$$
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$$90^o$$
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
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$$1$$
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$$4/3$$
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$$3/4$$
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$$4/5$$
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 ; $$\cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta $$
$$= \dfrac{1}{3}[(l+m+n)^{2}+(-l+m+n)^{2}+(l-m+n)^{2}+(l+m-n)^{2}]$$
$$=\dfrac{1}{3}[4l^{2}+4m^{2}+4n^{2}]=\dfrac{4}{3}(l^{2}+m^{2}+n^{2})$$
$$[\because l^{2}+m^{2}+n^{2}=1]=\dfrac{4}{3}$$
$$\therefore \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta =\dfrac{4}{3}$$
The direction ratios of the line, given by the planes x - y + z - 5 = 0, x - 3y - 6 = 0 are
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(3, 1, -2)
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(2, -4, 1)
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(1,-1, 1)
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(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 $$3,1,-2$$
If $$\overline { O A } = 3 \overline { i } + \overline { j } - \overline { k }$$, $$| \overline { A B } | = 2 \sqrt { 6 }$$ and AB has the direction ratios 1, -1 , 2 then $$| O B | =$$
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$$\sqrt { 35 }$$
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$$\sqrt { 41 }$$
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$$\sqrt { 26 }$$
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$$\sqrt { 55 }$$
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
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$$4i+j+3k$$
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$$4i+5j+3k$$
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4i-5j-3k
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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 $$\beta$$.
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$$\alpha = 1 , \beta = 2$$
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$$\alpha = 2 , \beta = 1$$
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$$\alpha = 1 , \beta = 1$$
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$$\alpha = 2 , \beta = 2$$
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 $$\begin{vmatrix} { l }_{ 1 } & m_{ 1 } & n_{ 1 } \\ { l }_{ 2 } & m_{ 2 } & n_{ 2 } \\ { l }_{ 3 } & m_{ 3 } & n_{ 3 } \end{vmatrix}$$
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$$0$$
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$$\pm 1$$
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$$\pm 2$$
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$$\pm \dfrac{1}{2}$$
The point where $$\vec{ x }$$ which is perpendicular to $$(2,-3,1)$$ and $$(1,-2,3)$$ and which satisfies the condition $$\vec { x } \cdot ( \hat { i } + 2 \hat { j } - 7 \hat{ k } ) = 10$$
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$$\left(3,5,1\right)$$
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$$\left(7,-5,1\right)$$
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$$\left(3,-5,1\right)$$
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$$\left(7,5,1\right)$$
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
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$$x+y+z=6$$
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$$x-y+z=6$$
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$$x-y-z=6$$
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$$x+y+z+6=0$$
If $$A(p,q,r)$$ 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
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$$7$$
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$$12$$
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$$5$$
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$$None$$ $$of$$ $$these$$
The angle between the lines whose de's satisfy the equation $$l+m+m=0$$ and $$l^2+m^{2}-n^{2}=0$$ is
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$$\dfrac{\pi}{6}$$
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$$\dfrac{\pi}{2}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{4}$$
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.
The angle between the lines, whose direction ratios are $$1,1,2$$ and $$\sqrt { 3 } - 1 , - \sqrt { 3 } - 1,4 ,$$ is
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$${45} ^ { \circ }$$
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$${30} ^ { \circ }$$
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$${60} ^ { \circ }$$
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$${90} ^ { \circ }$$
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:
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$$\dfrac { 1 }{ \sqrt { 35 } } ,-\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } } $$
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$$-\dfrac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } \dfrac { 3 }{ \sqrt { 35 } } $$
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$$\dfrac { 1 }{ \sqrt { 35 } } ,\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } } $$
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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
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$$\dfrac{3}{4}$$
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$$\dfrac{4}{3}$$
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$$\dfrac{7}{5}$$
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$$1$$
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
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$$l_{1}+ l_{2}+ l_{3},m_{1}+m_{2}+m_{3},n_{1}+n_{2}+n_{3}$$
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$$\dfrac{l_{1}+l_{2}+l_{3}}{\sqrt{3}},\dfrac{m_{1}+m_{2}+m_{3}}{\sqrt{3}},\dfrac{n_{1}+n_{2}+n_{3}}{\sqrt{3}}$$
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$$\dfrac{l_{1}+l_{2}+l_{3}}{3},\dfrac{m_{1}+m_{2}+m_{3}}{3},\dfrac{n_{1}+n_{2}+n_{3}}{3}$$
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$$None\ of\ these$$
$$l=m=n=1$$ represent the direction cosines of the
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$$x-$$ axis
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$$y-$$ axis
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$$z-$$ axis
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$$none\ of\ these$$
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
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-1
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1
0%
2
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0
Explanation
If the area of triangle is zero, then the points are collinear.
Points are collinear.
Point are $$(p, o)(o, q)(i,q)$$
$$\Delta =\dfrac {1}{2}[p(q-1)-0(1-0)+1(0-q)]$$
$$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$$
$$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$$
$$\Rightarrow \dfrac{1}{2}[pq-(p+q)]=o$$
$$\Rightarrow pq=p+q\Rightarrow \dfrac {p+q}{pq}=1$$
$$\Rightarrow \dfrac{1}{p}+\dfrac{1}{q}=1$$
The direction ratios of the line
$$x-y+z-5=\quad 0\quad =\quad x-3y-6\quad are$$
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$$3,1,-2$$
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$$2,-4,1$$
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$$\frac { 3 }{ \sqrt { 14 } } ,\frac { 1 }{ \sqrt { 14 } } ,\frac { -2 }{ \sqrt { 14 } } $$
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$$\frac { 2 }{ \sqrt { 41 } } ,\frac { -4 }{ \sqrt { 41 } } ,\frac { 1 }{ \sqrt { 41 } } $$
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
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$$l_1$$+$$l_2$$+$$l_3$$, $$m_1$$+$$m_2$$+$$m_3$$, $$n_1$$+$$n_2$$+$$n_3$$
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$$\dfrac{l_1+l_2+l_3}{\sqrt{3}} $$, $$\dfrac{m_1+m_2+m_3}{\sqrt{3}} $$, $$\dfrac{n_1+n_2+n_3}{\sqrt{3}} $$
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$$\dfrac{l_1+l_2+l_3}{3} $$, $$\dfrac{m_1+m_2+m_3}{3} $$, $$\dfrac{n_1+n_2+n_3}{3} $$
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None of these
If $$A(3\hat { i } +2\hat { j } +3\hat { k } ),B(-\hat { i } -\hat { j } +8\hat { k } ),C(-4\hat { i } +4\hat { j } +6\hat { k } )$$ are the vertices of a triangle then the equation of the line passing through the circumcentre and parallel to $$\vec { A B }$$ is
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$$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$
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$$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$
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$$\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } -\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$
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$$\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } -\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )$$
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
$$\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}$$
If $$\cos { \alpha ,\quad \cos { \beta ,\quad \cos { \gamma } } }$$ 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:
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$$<3,10,-7>$$
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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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$$
(0,1,1)
$$
0%
$$
(2,0,2)
$$
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$$
(1,0,0)
$$
0%
$$
(0,1,0)
$$
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 $$(0,1,1)$$
0:0:1
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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