Question

A plane Δ passes through the point P(3,−2,1) and is perpendicular to the vector v¯=4i^+7j^−4k^. The distance between Δ and the point Q(2,2,2) equals _____.​

Answer 1

Step-by-step explanation:

this is the answer for the question

Answer 2

Therefore the distance between plane and the point Q(2,2,2)is$=\frac{20}{9}$ units

Step-by-step explanation:

Given ,a plane passes through the point P(3,-2,1) and is perpendicular to the vector$\vec{v}=4 \hat{i}+7 \hat j-4\hat k$

Since the plane is perpendicular to the given vector then the direction ratio of the plane is (4,7,-4)

The equation of the plane whose direction ratio is (a,b,c) and passes through the point(x₁,y₁,z₁) is

a(x-x₁)+b(y-y₁)+c(z-z₁)=0

Therefore the equation of required plane is

4(x-3)+7(y+2)-4(z-1)=0

⇔4x-12+7y+14-4z+4=0

⇔4x+7y-4z+6=0

Therefore the distance between plane and the point Q(2,2,2)is

$=|\frac{4.2+7.2-4.2+6}{\sqrt{4^2+7^2+(-4)^2} } |$

$=|\frac{20}{\sqrt{81} } |$

$=\frac{20}{9}$  units