The plane passing through the point whose position vector is i + j − k and parallel to the lines 1 x = 2 y = 3 z and − 1 x − 1 = − 2 y + 1 = 1 z − 1 has l , m , n as direction cosines of i

Question

The plane passing through the point whose position vector is $i + j - k$ and parallel to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x - 1}{- 1} = \frac{y + 1}{- 2} = \frac{z - 1}{1}$ has $l, m, n$ as direction cosines of its normal then $∣ l + m + n ∣ =$

Options:

Answer:

C

Solution:

$a + 2 b + 3 c = 0$

$- a - 2 b + c = 0$

$\Rightarrow a : b : c = 2 : - 1 : 0$

TestHub · Accepted Answer

a + 2 b + 3 c = 0 − a − 2 b + c = 0 ⇒ a : b : c = 2 : − 1 : 0

Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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