Find the distance of point -2 -3 -4 k- from the line r -2 -k-3 -9 k- measured parallel to the plane x - y -2 z -3-0-

The line r⃗ = î + 2 ĵ k̂ + λ (î + 3 ĵ 9 k̂) can be rewritten as x 1/1 = y 2/3 = z+1/ 9 = λ. So coordinates of the random point on this line : P(λ + 1, 3 λ + ...

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Standard XII

Mathematics

Family of Planes Passing through the Intersection of Two Planes

Find the dist...

Question

Find the distance of point $- 2^i + 3^j - 4^k$ from the line $\to r =^i + 2^j -^k + λ (^i + 3^j - 9^k)$ measured parallel to the plane x - y + 2z - 3 = 0.

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Solution

The line $\to r =^i + 2^j -^k + λ (^i + 3^j - 9^k)$ can be rewritten as $\frac{x - 1}{1} = \frac{y - 2}{3} = \frac{z + 1}{- 9} = λ .$
So coordinates of the random point on this line : $P (λ + 1, 3 λ + 2, - 9 λ - 1) .$
Let A(-2, 3, -4) be the point whose position vector is $- 2^i + 3^j - 4^k$ .
The d.r.'s of the line AP parallel to the plane x - y + 2z - 3 = 0 are $λ + 3, 3 λ - 1, - 9 λ + 3.$
As normal to the plane shall be $⊥$ to line AP so, $1 (λ + 3) - (3 λ - 1) + 2 (3 - 9 λ) = 0 \Rightarrow λ = \frac{1}{2} .$ So $P (\frac{3}{2}, \frac{7}{2}, - \frac{11}{2}) .$

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Find the distance of point −2^i+3^j−4^k from the line →r=^i+2^j−^k+λ(^i+3^j−9^k) measured parallel to the plane x - y + 2z - 3 = 0.

Find the distance of point $- 2^i + 3^j - 4^k$ from the line $\to r =^i + 2^j -^k + λ (^i + 3^j - 9^k)$ measured parallel to the plane x - y + 2z - 3 = 0.