MCQ Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 10

The direction cosine of a line equally inclined to the axes are

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$$\displaystyle \frac { 1 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 1 }{ 3 } $$
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$$\displaystyle -\frac { 1 }{ 3 } ,-\frac { 1 }{ 3 } ,-\frac { 1 }{ 3 } $$
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$$\displaystyle \frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } ,\frac { 1 }{ \sqrt { 3 } } $$
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none of these

If $$l_1, m_1, n_1$$ and $$l_2, m_2, n_2$$ be the DC's of two concurrent lines, the direction cosines of the line bisecting the angles between them are proportional to

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$$l_1 l_2, m_1 m_2, n_1, n_2$$
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$$l_1m_2, l_1 n_2, l_1 n_3$$
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$$l_1 + l_2, m_1 + m_2, n_1 + n_2$$
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None of these

The angle between the lines $$\displaystyle 2x=3y=-z\:and\:6x=-y=-4z$$ is

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$$\displaystyle \frac{\pi}{2}$$
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$$0$$
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$$\displaystyle \frac{\pi}{6}$$
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$$\displaystyle \frac{\pi}{4}$$

The direction cosines of a line whose equations are $$\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-2}{-3}$$

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$$\dfrac{1}{\sqrt{14}},\dfrac{-3}{\sqrt{14}},\dfrac{2}{\sqrt{14}}$$
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$$\dfrac{2}{\sqrt{29}},\dfrac{4}{\sqrt{29}},\dfrac{-3}{\sqrt{29}}$$
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$$\dfrac{1}{\sqrt{29}},\dfrac{-3}{\sqrt{29}},\dfrac{2}{\sqrt{29}}$$
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$$2,4,-3$$

If the foots of the perpendicular from the origin to a plane is $$(a,b,c)$$, the equation of the plane is

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$$\displaystyle \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =3$$
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$$ax+by+cz=3$$
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$$ax+by+cz={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$$
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$$ax+by+cz=a+b+c$$

If equation of the plane through the straight line $$\displaystyle \dfrac{x -1}{2}=\dfrac{y +2}{-3}=\dfrac{z}{5}$$ and perpendicular to the plane $$x - y + z + 2 = 0 \: $$is$$ \:ax- by + cz + 4 = 0,$$ then find the value of $$ a^2 + b^2 + c$$

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$$12$$
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$$14$$
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$$16$$
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$$18$$

if a line makes angles $$\alpha,\beta,\gamma,\delta$$ with four diagonals a cube then value of $$sin^{2}\alpha+sin^{2}\beta+sin^{2}\gamma+sin^{2}\delta$$ equals

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$$2$$
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$$\displaystyle \frac{4}{3}$$
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$$\displaystyle \frac{8}{3}$$
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$$1$$

Three points whose position vectors are $$\overrightarrow{a}$$, $$\overrightarrow{b}$$, $$\overrightarrow{c}$$ will be collinear if

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$$\lambda \overrightarrow{a}+\mu \overrightarrow{b}=\left ( \lambda +\mu \right )\overrightarrow{c}$$
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$$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}$$
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$$\begin{bmatrix}

\overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c}

\end{bmatrix}=0$$
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None of these

Determine the equation of the plane on which the co - ordinates of the foot of perpendicular drawn from origin O is the point $$\displaystyle P\left ( \alpha ,\beta ,\gamma \right ).$$

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$$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha ^{3}+\beta ^{3}+\gamma ^{3}$$
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$$\displaystyle \alpha x+\beta y+\gamma z=\alpha ^{2}+\beta ^{2}+\gamma ^{2}$$
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$$\displaystyle \alpha^{2} x+\beta^{2} y+\gamma^{2} z=\alpha +\beta +\gamma $$
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none of these

If $$\displaystyle l_{1},m_{1},n_{1}$$ and $$\displaystyle l_{2},m_{2},n_{2}$$ are D.C.'s of the two lines inclined to each other at an angle $$\displaystyle \theta $$, then the D. C.'s of the internal and external bisectors of the angle between these lines are

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$$\displaystyle \frac{l_{1}+l_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{m_{1}+m_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{n_{1}+n_{2}}{2\sin \left ( \theta /2 \right )}$$
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$$\displaystyle \frac{l_{1}+l_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{m_{1}+m_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{n_{1}+n_{2}}{2\cos \left ( \theta /2 \right )}$$
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$$\displaystyle \frac{l_{1}-l_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{m_{1}-m_{2}}{2\sin \left ( \theta /2 \right )},\, \frac{n_{1}-n_{2}}{2\sin \left ( \theta /2 \right )}$$
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$$\displaystyle \frac{l_{1}-l_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{m_{1}-m_{2}}{2\cos \left ( \theta /2 \right )},\, \frac{n_{1}-n_{2}}{2\cos \left ( \theta /2 \right )}$$

Explanation

If $$l_1,\; m_1, n_1$$ and $$l_2,\; m_2, n_2$$ are D.C.s of two lines then,

$$l_1l_2+m_1m_2+n_1n_2= \cos \theta$$

The direction ratios of the internal angle bisector are $$l_1+l_2,\; m_1+m_2, \; n_1+n_2$$.

Hence, the direction cosines of the internal angle bisector are $$ \displaystyle \dfrac { { l }_{ 1 }+{ l }_{ 2 } }{ \sqrt { \sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } } } ,\dfrac { { m }_{ 1 }+{ m }_{ 2 } }{ \sqrt { \sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } } } ,\dfrac { { n }_{ 1 }+{ n }_{ 2 } }{ \sqrt { \sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } } } $$

where, $$\sum { { \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 } } \\ ={ \left( { l }_{ 1 }+{ l }_{ 2 } \right) }^{ 2 }+{ \left( { m }_{ 1 }+{ m }_{ 2 } \right) }^{ 2 }+{ \left( { n }_{ 1 }+{ n }_{ 2 } \right) }^{ 2 }\\ =\left( { { l }_{ 1 } }^{ 2 }+{ { m }_{ 1 } }^{ 2 }+{ { n }_{ 1 } }^{ 2 } \right) +\left( { { l }_{ 2 } }^{ 2 }+{ { m }_{ 2 } }^{ 2 }+{ { n }_{ 2 } }^{ 2 } \right) +2\left( { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 } \right) \\ =1+1+2\cos { \theta } =2+2\cos { \theta } =4\cos^{2} { \dfrac { \theta }{ 2 } } $$

Hence direction cosines of internal angle bisector are $$ \displaystyle \dfrac { { l }_{ 1 }+{ l }_{ 2 } }{ 2\cos { \dfrac { \theta }{ 2 } } } ,\dfrac { { m }_{ 1 }+{ m }_{ 2 } }{ 2\cos { \dfrac { \theta }{ 2 } } } ,\dfrac { { n }_{ 1 }+{ n }_{ 2 } }{ 2\cos { \dfrac { \theta }{ 2 } } } $$

Similarly , D.R.s of external angle bisector are $${ l }_{ 1 }-{ l }_{ 2 },\; { m }_{ 1 }-{ m }_{ 2 },\; { n }_{ 1 }-{ n }_{ 2 }$$

So, the direction cosines of the external angle bisector are $$ \displaystyle \dfrac { { l }_{ 1 }-{ l }_{ 2 } }{ \sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } } ,\dfrac { { m }_{ 1 }-{ m }_{ 2 } }{ \sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } } ,\dfrac { { n }_{ 1 }-{ n }_{ 2 } }{ \sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } } $$

where, $$\sum { { \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 } } \\ ={ \left( { l }_{ 1 }-{ l }_{ 2 } \right) }^{ 2 }+{ \left( { m }_{ 1 }-{ m }_{ 2 } \right) }^{ 2 }+{ \left( { n }_{ 1 }-{ n }_{ 2 } \right) }^{ 2 }\\ =\left( { { l }_{ 1 } }^{ 2 }+{ { m }_{ 1 } }^{ 2 }+{ { n }_{ 1 } }^{ 2 } \right) +\left( { { l }_{ 2 } }^{ 2 }+{ { m }_{ 2 } }^{ 2 }+{ { n }_{ 2 } }^{ 2 } \right) -2\left( { l }_{ 1 }{ l }_{ 2 }+{ m }_{ 1 }{ m }_{ 2 }+{ n }_{ 1 }{ n }_{ 2 } \right) \\ =1+1-2\cos { \theta } =2-2\cos { \theta } =4\sin ^{ 2 }{ \dfrac { \theta }{ 2 } } $$.

Hence, direction cosines of external angle bisector are $$ \displaystyle \dfrac { { l }_{ 1 }-{ l }_{ 2 } }{ 2\sin { \dfrac { \theta }{ 2 } } } ,\dfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 2\sin { \dfrac { \theta }{ 2 } } } ,\dfrac { { n }_{ 1 }-{ n }_{ 2 } }{ 2\sin { \dfrac { \theta }{ 2 } } } $$

Hence, options 'B' and 'D' are correct.

$$O$$ is the origin and $$A$$ is the point $$\displaystyle \left ( a,b,c \right ).$$ Find the direction cosines of the join of $$OA$$ and deduce the equation of the plane through $$A$$ at right angles to $$OA$$.

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$$\displaystyle ax+by+cz=a^{2}-b^{2}-c^{2}$$
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$$\displaystyle ax+by+cz=a^{2}+b^{2}+c^{2}$$
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$$\displaystyle ax-by-cz=a^{2}+b^{2}+c^{2}$$
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$$\displaystyle ax-by-cz=a^{2}-b^{2}-c^{2}$$

The direction cosines of the lines bisecting the internal angle $$\theta$$ between the lines whose direction cosines are $$l_{1},m_{1},n_{1}$$ and $$l_{2},m_{2},n_{2}$$ are

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$$< l_{1}+l_{2},m_{1} +m_{2},n_{1}+n_{2}> $$
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$$\displaystyle < \frac{l_{1}+l_{2}}{2\sin \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\sin \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\sin \frac{\theta}{2}}>$$
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$$\displaystyle < \frac{l_{1}+l_{2}}{2\cos \frac{\theta}{2}},\frac{m_{1}+m_{2}}{2\cos \frac{\theta}{2}},\frac{n_{1}+n_{2}}{2\cos \frac{\theta}{2}}>$$
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none of these

The projection of a directed line segment on the co-ordinate axes are $$12, 4, 3$$, the DC's of the line are

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$$\displaystyle \frac {-12}{13}, \frac {-4}{13}, \frac {-3}{13}$$
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$$\displaystyle \frac {12}{13}, \frac {4}{13}, \frac {3}{13}$$
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$$\displaystyle \frac {12}{13}, \frac {-4}{13}, \frac {3}{13}$$
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$$\displaystyle \frac {12}{13}, \frac {4}{13}, \frac {-3}{13}$$

The projections of a line segment on $$x, y, z$$ axes are $$12, 4, 3$$. The length and the direction cosines of the line segments are

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$$\displaystyle 13,< 12/13, 4/13, 3/13> $$
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$$\displaystyle 19,< 12/19, 4/19, 19> $$
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$$\displaystyle 11,< 12/11, 4/11, 3/11> $$
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None of these

The line $$\displaystyle \frac{x - 1}{2} = \frac{y}{-1} = \frac{z + 2}{2}$$ cuts the plane $$\displaystyle x + y + z = 1$$ at $$\displaystyle P$$. If the foot of the perpendicular from $$\displaystyle P$$ to a point $$Q\displaystyle \left ( 3, \: -4, \: 1 \right )$$ on the plane $$S$$ then the equation of the plane $$S$$ is

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$$\displaystyle 3x - 2y - z = 0$$
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$$\displaystyle 2x - y + 2z = 12$$
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$$\displaystyle 2x - 10y + 5z = 51$$
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none of these

The projection of a line segment joining the points $$P\left ( x_{1},y_{1},z_{1}, \right )$$ and $$Q\left ( x_{1},y_{1},z_{1}, \right )$$ on another line whose DC's are $$l, m, n$$ is given by

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$$l\left ( x_{1}+x_{2} \right )+m\left (y _{2}+y_{2} \right )+n\left ( z_{1}+z_{2} \right )$$
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$$\displaystyle 2\left [ \frac{\left ( lx_{2}+my_{2}+nz_{2} \right )}{2}-\frac{\left (lx_{1}+my_{1}+nz_{1} \right )}{2} \right ]$$
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$$\displaystyle l\left ( x_{2}-x_{1} \right )+ m\left ( y_{2}-y_{1} \right )+n\left ( z_{2}-z_{1} \right )$$
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$$\displaystyle \frac{x_{2}-x_{1}}{l}+\frac{y_{2}-y_{1}}{m}+\frac{z_{2}-z_{1}}{n}$$

If $$A$$ , $$B$$ and $$C$$ are three collinear points, where $$A= i + 8 j - 5k $$, $$ B = 6i-2j$$ and $$C= 9i + 4j - 3 k$$, then $$B$$ divides $$AC$$ in the ratio of :

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$$\dfrac{5}{7}$$
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$$\dfrac{5}{3}$$
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$$\dfrac{2}{3}$$
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None of these

If the position vectors of the points $$A$$, $$B$$, and $$C$$ be $$i + j $$ , $$i - j$$ and $$ai + bj+ ck$$ respective;y , then the points $$A$$, $$B$$ and $$C$$ are collinear if:

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$$a = b = c = 1$$
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$$a= 1$$ , $$b$$ and $$c$$ are arbitrary scalars
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$$a =b= c= 0$$
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$$c = 0$$ , $$a =1$$ and $$b$$ is a arbitrary scalar.

The angle between the lines whose direction cosines are given by the equations $${l}^{2}+{m}^{2}-{n}^{2}=0,l+m+n=0$$ is

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$$\displaystyle\frac{\pi}{6}$$
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$$\displaystyle\frac{\pi}{4}$$
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$$\displaystyle\frac{\pi}{3}$$
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$$\displaystyle\frac{\pi}{2}$$

If direction cosines of two lines are proportional to $$(2,3,-6)$$ and $$(3,-4,5)$$, then the acute angle between them is

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$$\cos ^{ -1 }{ \left( \cfrac { 49 }{ 36 } \right) } $$
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$$\cos ^{ -1 }{ \left( \cfrac { 18\sqrt { 2 } }{ 35 } \right) } $$
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$${96}^{o}$$
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$$\cos ^{ -1 }{ \left( \cfrac { 18 }{ 35 } \right) } $$

If the median through $$A$$ of a $$\triangle ABC$$ having vertices $$A\equiv \left( 2,3,5 \right),$$ $$B\equiv \left( -1,3,2 \right) $$ and $$C\equiv \left( \lambda ,5,\mu \right) $$ is equally inclined to the axes, then

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$$\lambda =7$$
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$$\mu=10$$
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$$\lambda =10$$
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$$\mu=7$$

Find the unit vectors perpendicular to the following pair of vectors:

$$2i+j+k$$, $$i-2j+k$$

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$$\displaystyle \frac{1}{\sqrt{35}}(3i-j-5k)$$
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$$\displaystyle \frac{1}{\sqrt{35}}(3i+j-5k)$$
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$$\displaystyle \frac{1}{\sqrt{27}}(i-j-5k)$$
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$$\displaystyle \frac{1}{\sqrt{27}}(i-j+5k)$$

If the direction cosines of two lines are given by $$l+m+n=0$$ and $$l^2-5m^2+n^2=0$$, then the angle between them is

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

The vector equation of the line $$\displaystyle L_{1}$$ is $$\displaystyle a+\lambda \overline{b}$$ then $$\displaystyle \overline{a}$$ equals

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

The, position vector of the foot of the $$\perp$$er drawn from origin to the plane is $$\displaystyle 4\hat{i}-2\hat{j}-5\hat{k} $$ then equation of the plane is

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$$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}+5\hat{k} \right )=45$$
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$$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )=45$$
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$$\displaystyle \overline{r}.\left ( 4\hat{i}-2\hat{j}-5\hat{k} \right )+45=0$$
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$$\displaystyle \overline{r}.\left ( 4\hat{i}+2\hat{j}-5\hat{k} \right )=37$$

If the points $$a(cos \alpha + i sin \alpha)$$ , $$b(cos \beta + i sin \beta)$$ and $$c(cos \gamma + isin \gamma)$$ are collinear then the value of $$|z|$$ is:
( where $${z = bc \ sin(\beta-\gamma) + ca \ sin(\gamma-\alpha) + ab \ sin(\alpha - \beta) + 3i -4k}$$ )

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$$2$$
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$$5$$
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$$1$$
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None of these.

The angle between two diagonals of a cube is

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$$\cos^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$$
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$$\cos^{-1}\left (\dfrac {1}{3}\right )$$
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$$30^{\circ}$$
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$$45^{\circ}$$

Equation of the plane through the mid-point of the line segment joining the points $$P(4, 5, -10), \,Q(-1, 2, 1)$$ and perpendicular to $$PQ$$ is

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$$r. \left( \dfrac{}3{}2 \widehat i + \dfrac{7}{2} \widehat j - \dfrac{9}{2} \widehat k\right ) =45$$
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$$r. (- \widehat i + 2 \widehat j + \widehat k) = \dfrac{135}{2}$$
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$$r. (5 \widehat i + 3 \widehat j - 11 \widehat k) + \dfrac{135}{2} = 0$$
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$$r. (4 \widehat i + 5 \widehat j - 10 \widehat k) = 85$$
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$$r. (5\widehat i + 3 \widehat j - 11\widehat k) = \dfrac{135}{2}$$

If $$(2, -1, 3)$$ is the foot of the perpendicular drawn from the origin to the plane, then the equation of the plane is

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$$2x + y - 3z + 6 = 0$$
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$$2x - y + 3z - 14 = 0$$
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$$2x - y + 3z - 13 = 0$$
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None of these

The equation of the line parallel to $$\cfrac { x-3 }{ 1 } =\cfrac { y+3 }{ 5 } =\cfrac { 2z-5 }{ 3 } $$ and passing through the point $$(1,3,5)$$ in vector form, is:

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$$\vec { r } =\left( \vec { i } +5\vec { j } +3\vec { k } \right) +t\left( \vec { i } +3\vec { j } +5\vec { k } \right) $$
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$$\vec { r } =\left( \vec { i } +3\vec { j } +5\vec { k } \right) +t\left( \vec { i } +5\vec { j } +3\vec { k } \right) $$
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$$\vec { r } =\left( \vec { i } +5\vec { j } +\cfrac { 3 }{ 2 } \vec { k } \right) +t\left( \vec { i } +3\vec { j } +5\vec { k } \right) $$
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$$\vec { r } =\left( \vec { i } +3\vec { j } +5\vec { k } \right) +t\left( \vec { i } +5\vec { j } +\cfrac { 3 }{ 2 } \vec { k } \right) $$

What is the angle between the lines $$\cfrac { x-2 }{ 1 } =\cfrac { y+1 }{ -2 } =\cfrac { z+2 }{ 1 } $$ and $$\cfrac { x-1 }{ 1 } =\cfrac { 2y+3 }{ 3 } =\cfrac { z+5 }{ 2 } =?$$

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$$\cfrac { \pi }{ 2 } $$
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$$\cfrac { \pi }{ 3 } $$
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$$\cfrac { \pi }{ 6 } $$
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None of the above

A plane mirror is placed at the origin so that the direction ratios of its normal are $$(1,-1,1)$$. A ray of light, coming along the positive direction of the x-axis, strikes the mirror. The direction cosines of the reflected ray are

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

The equation of the plane passing through (1, -2, 4), (3, -4, 5) and perpendicular to yz-plane is.

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2y + z = 0
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y + 2y + 6 = 0
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y + 2y - 6 = 0
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3y + 2z - 2 = 0

$$L_{1}$$ and $$L_{2}$$ are two lines whose vector equations are
$$L_{1} = \vec {r} = \lambda [(\cos \theta + \sqrt {3})\hat {i} + (\sqrt {2}\sin \theta)\hat {j} + (\cos \theta - \sqrt {3})\hat {k}]$$
$$L_{2} = \vec {r} = \mu (a\hat {i} + b\hat {j} + c\hat {k})$$, where $$\lambda$$ and $$\mu$$ are scalars and $$\alpha$$ is the acute angle between $$L_{1}$$ and $$L_{2}$$. If the angle $$'\alpha'$$ is independent of $$\theta$$ then the value of $$'\alpha'$$ is

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$$\dfrac {\pi}{6}$$
0%
$$\dfrac {\pi}{4}$$
0%
$$\dfrac {\pi}{3}$$
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None of these

If points $$P\left( 4,5,x \right) ,Q\left( 3,y,4 \right) $$ and $$ R\left( 5,8,0 \right) $$ are colinear, then the value of $$x+y$$ is

0%
$$-4$$
0%
$$3$$
0%
$$5$$
0%
$$4$$

If $$\alpha,\beta,\gamma\in[0,2\pi]$$, then the sum of all possible values of $$\alpha, \beta,\gamma$$ if $$\sin \alpha=-\dfrac{1}{\sqrt{2}}$$, $$\cos \beta=-\dfrac{1}{2}$$, $$\tan \gamma=-\sqrt{3}$$, is

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$$\dfrac{22\pi}{3}$$
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$$\dfrac{21\pi}{3}$$
0%
$$\dfrac{20\pi}{3}$$
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$$8\pi$$

$$If\,\ell ,\,m,\,n\,\& \,\ell ',\,m',\,n'\,$$ be the cosine of two lines which include then

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$$\cos \,\theta = \ell \ell ' + mm' + nn'$$
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$$\sin \,\theta = \ell \ell ' + mm' + nn'$$
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$$\cos \,\theta \, = mn' + m'n + n\ell ' + n'\ell + \ell m' + \ell 'm$$
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$$\sin \,\theta \, = mn' + m'n + n\ell ' + n'\ell + \ell m' + \ell 'm$$

In $$\Delta ABC, |\bar{CB}| = a, |\bar{CA}| = b, |\bar{AB}| = c$$. $$CD$$ is median through the vertex $$C$$. Then $$\bar{CA}.\bar{CD}$$ equals.

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$$\dfrac{1}{4}(3a^2 + b^2 - c^2)$$
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$$\dfrac{1}{4}(a^2 + 3b^2 - c^2)$$
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$$\dfrac{1}{4}(a^2 + b^2 - 3c^2)$$
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$$\dfrac{1}{4}(-3a^2 + b^2 - c^2)$$

The equation of plane passing through a point $$A(2, - 1, 3)$$ and parallel to the vectors $$a= (3, 0, - 1)$$ and $$b=(- 3, 2, 2)$$ is:

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$$2x - 3y + 6z - 25 = 0$$
0%
$$2x - 3y + 6z + 25 = 0$$
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$$3x - 2y + 6z - 25 = 0$$
0%
$$3x - 2y + 6z + 25 = 0$$

If a = 4i + 3j and b be two vectors perpendicular to each other on the xy- plane. Then, a vector in the same plane having projections 1 and 2 along a and b respectively, is

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i + 2j
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2i - j
0%
2i + j
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None of these

If a line makes angle $$90^{o},135^{o},45^{o}$$ with the $$X-$$,$$Y-$$ and $$Z-$$axes respectively, then its direction cosines are

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$$0,\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}$$
0%
$$0,-\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}$$
0%
$$0,\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}$$
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$$0,-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}$$

In the isosceles $$\triangle$$ABC, $$|AB| = |BC|=8$$ and point E divides AB internally in the ratio 1 : 3 then the cosine of angel between CE and CA is (where, |CA| = 12) ?

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$$-\dfrac { \sqrt [ 3 ]{ 7 } }{ 8 }$$
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$$\dfrac { \sqrt [ 3 ]{ 8 } }{ 17 }$$
0%
$$\dfrac { \sqrt [ 3 ]{ 7 } }{ 8 }$$
0%
$$-\dfrac { \sqrt [ 3 ]{ 8 } }{ 17 }$$

The direction cosines of the line $$x=44z+3; y=2-2\sqrt{19}z$$ is...

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$$\dfrac{44}{\sqrt{2013}};\dfrac{-2\sqrt{19}}{\sqrt{2013}};\dfrac{1}{\sqrt{2013}}$$
0%
$$\dfrac{-44}{\sqrt{2013}};\dfrac{2\sqrt{19}}{\sqrt{2013}};\dfrac{1}{\sqrt{2013}}$$
0%
$$\dfrac{44}{\sqrt{2013}};\dfrac{2\sqrt{19}}{\sqrt{2013}};\dfrac{1}{\sqrt{2013}}$$
0%
$$\dfrac{-44}{\sqrt{2013}};\dfrac{-2\sqrt{19}}{\sqrt{2013}};\dfrac{-1}{\sqrt{2013}}$$

The distance of the point $$3\hat {i}+5\hat {k}$$ from the line parallel to $$6\hat {i}+\hat {j} 2\hat {k}$$ and passing through the point $$8\hat {i}+3\hat {j}+\hat {k}$$ is

0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$

Three points whose position vectors are $$x\bar{i}+y\bar{j}+z\bar{k}$$, $$\bar{i}+2\bar{j}$$ and $$-\bar{i}-\bar{j}$$ are collinear, then relation between $$x, y, z$$ is?

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$$x-2y=1, z=0$$
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$$z+y=1, z=0$$
0%
$$x-y=1, z=0$$
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None of these

A line makes equal angles with the coordinate axis. The direction cosines of this line are

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$$\left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)$$
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$$\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)$$
0%
$$\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{3},\dfrac{1}{3}\right)$$
0%
$$\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)$$

If vector $$\overrightarrow{a}+\overrightarrow{b}$$ bisects the between $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$, then $$\overrightarrow|{a}|=\overrightarrow|{b}|$$.

0%
True
0%
False

The direction Ratio's of normal of the plane through $$(1, 0, 0), (0, 1, 0)$$ which makes angle $$\pi/4$$ with plane $$x+y =3$$ are

0%
$$1, \sqrt{2}, 1$$
0%
$$1, \sqrt{2}, \sqrt{2}$$
0%
$$ \sqrt{2}, 1, 1$$
0%
$$1, 1, \sqrt{2}$$

A line passes through the point $$(6, -7, -1)$$ and $$(2, -3, 1)$$. Then the sum of the direction cosines of the line, if the line makes acute angle with positive direction of x-axis, is?

0%
$$1/3$$
0%
$$4/3$$
0%
$$-1/3$$
0%
$$2/3$$

The direction ratios of a line perpendicular to both the lines whose direction ratios are $$3,-2,4$$ and $$1,3,-2$$

0%
$$2,-5,6$$
0%
$$4,-10,12$$
0%
$$-8,10,11$$
0%
$$-8,10,-11$$

CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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