Question

Given that p=3i+j+2kand q=i-2j-4k are the position vectors

Answer 1

The equation for the plane passing through Q and perpendicular to line PQ is r (2 i + 3 j + 6 k) + 28 = 0.

Let position vector of point P be:

p = 3 i + j + 2 k

Let position vector of point Q be:

q = i - 2 j - 4 k

So, PQ = Q - P

PQ = n = i - 2 j - 4 k - (3 i + j + 2 k)

n = i - 2 j - 4 k - 3 i - j - 2 k

n = - 2 i - 3 j - 6 k

The Equation of plane passing through point Q and perpendicular to PQ will be:

(r - q).n = 0

r n = q n

q n = (i - 2 j - 4 k) . (- 2 i - 3 j - 6 k)

q n = - 2 + 6 + 24

q n = 28

r n = 28

r (- 2 i - 3 j - 6 k) = 28

r (2 i + 3 j + 6 k) + 28 = 0

Therefore the equation for the plane passing through Q and perpendicular to line PQ is r (2 i + 3 j + 6 k) + 28 = 0.

Learn more equation for the plane here:

brainly.com/question/18831322

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