Given -3-2k- and -2-4k- are the position vectors of the points A and B- Then the distance of the point -k- from the plane passing through B and perpendicular to AB is 27 - k -Find sum of digits of k -

The equation of plane is (r⃗ β⃗).(α⃗ β⃗)=0 ⇒ 2x y+6z+24=0 Therefore, perpendicular distance from ( 1,1,1) is d=| 2 1+6+24 √( 4+1+36 ) n ...

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Perpendicular Distance of a Point from a Plane

Given α⃗=3î...

Question

Given $\to α = 3^i +^j + 2^k$ and $\to β =^i + 2^j - 4^k$ are the position vectors of the points A and B. Then the distance of the point $-^i +^j +^k$ from the plane passing through B and perpendicular to AB is $\frac{27}{\sqrt{k}}$ .
Find sum of digits of k .

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\to α = 3^i +^j + 2^k, \to β =^i - 2^j - 4^k

A

B

-^i +^j +^k

B

A B

\to α = 3^i +^j + 2^k

\to β =^i - 2^j - 4^k

A

B .

-^i +^j +^k

B

A B

\to α = 3^i +^j + 2^k

\to β =^i - 2^j - 4^k

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-^i +^j +^k

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\to a = 3^i +^j + 2^k, \to b =^i - 2^j - 4^k

-^i +^j +^k

^i +^j - 2^k, 2^i -^j +^k

^i + 2^j +^k

\to r = 3^i -^j -^k + λ (2^i - 2^j +^k)

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