Question

The direction cosines of two lines are related by $$l + m + n = 0$$    and $$a{l^2} + b{m^2} + c{n^2} = 0.$$    The lines are parallel if :

A. $$a + b + c = 0$$
B. $${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$$  
C. $$a = b = c$$
D. none of these
Answer :   $${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$$
Solution :
For $$n = - \left( {l + m} \right),$$    the second relation gives
$$\eqalign{ & a{l^2} + b{m^2} + c{\left( {l + m} \right)^2} = 0 \cr & {\text{or }}\left( {a + c} \right){l^2} + 2clm + \left( {b + c} \right){m^2} = 0 \cr} $$
For parallel lines, the two roots must be equal
$$\eqalign{ & \Rightarrow 4{c^2} - 4\left( {b + c} \right)\left( {a + c} \right) = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr} $$

Releted MCQ Question on
Geometry >> Three Dimensional Geometry

Releted Question 1

The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$     lies in the plane $$2x - 4y + z = 7,$$    is :

A. $$7$$
B. $$ - 7$$
C. no real value
D. $$4$$
Releted Question 2

If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$      and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$     intersect, then the value of $$k$$ is :

A. $$\frac{3}{2}$$
B. $$\frac{9}{2}$$
C. $$ - \frac{2}{9}$$
D. $$ - \frac{3}{2}$$
Releted Question 3

A plane which is perpendicular to two planes $$2x - 2y + z = 0$$    and $$x - y + 2z = 4,$$    passes through $$\left( {1,\, - 2,\,1} \right).$$   The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$  is :

A. $$0$$
B. $$1$$
C. $$\sqrt 2 $$
D. $$2\sqrt 2 $$
Releted Question 4

Let $$P\left( {3,\,2,\,6} \right)$$   be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$  is parallel to the plane $$x-4y+3z=1$$    is :

A. $$\frac{1}{4}$$
B. $$ - \frac{1}{4}$$
C. $$\frac{1}{8}$$
D. $$ - \frac{1}{8}$$

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