Question

A plane P passes through a point P (3, -2, 1) and is perpendicular to the vector

$\overrightarrow{V}=4\hat{i}+7\hat{j}-4\hat{k}.$ The distance between the plane P and the plane. $\overrightarrow{r}.(4\hat{i}+7\hat{j}-4\hat{k})+33=0$

• A. 3
• B. 2
• C. 1
• D. $\frac{28}{9}$

Answer 1

The correct option is A

3

Equation of the plane passing through the point P(3, -2, 1) is

A(x - 3) + B(y + 2) + C(z-1) = 0......(1)

equation (1) is perpendicular to $\overrightarrow{V}=4\hat{i}+7\hat{j}-4\hat{k}$

hence A = 4; B = 7 and C = - 4

hence equation of the plane is

4x + 7y – 4z + 6 = 0 ….(2)

given 4x + 7y – 4z + 33 = 0 ….(3)

$\therefore$ Distance between (2) and (3)

$=\left|\frac{27}{\sqrt{16+49+16}}\right|=\frac{27}{9}=3$