Planes in 3D

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$The value of m for which straight line 3 x - 2 y + z + 3 = 0 = 4 x - 3 y + 4 z + 1 is$ $parallel to the plane 2 x - y + m z - 2 = 0 is$

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Solution

$L e t {\bar{n}}_{1} = (3, - 2, 1), {\bar{n}}_{2} = (4, - 3, 4), {\bar{n}}_{3} = (2, - 1, m)$
$T h e n {\bar{n}}_{1} \times {\bar{n}}_{2} i s p a r a l l e l t o t h e l i n e a n d ⊥ t o n o r m a l {\bar{n}}_{3} t o t h e g i v e n p l a n e$
$\Rightarrow ({\bar{n}}_{1} \times {\bar{n}}_{2}) . {\bar{n}}_{3} = 0 o r [{\bar{n}}_{1} {\bar{n}}_{2} {\bar{n}}_{3}] = 0 \Rightarrow m = - 2$

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Planes in 3D

Remember concepts with our Masterclasses.

The value of m for which straight line 3x-2y+z+3=0=4x-3y+4z+1 is parallel to the plane 2x-y+mz-2=0 is

Let n¯1=3, −2, 1, n¯2=4, −3, 4, n¯3=2, −1, mThen n¯1×n¯2 is parallel to the line and ⊥ to normal n¯3 to the given plane⇒n¯1×n¯2.n¯3=0 or n¯1 n¯2 n¯3=0 ⇒m=−2

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$The value of m for which straight line 3 x - 2 y + z + 3 = 0 = 4 x - 3 y + 4 z + 1 is$ $parallel to the plane 2 x - y + m z - 2 = 0 is$