The distance of the plane through (2, 3, - 1) and at right angles to the vector 3 hati - 4 hatj + 7 hatk from the origin is

Question

The distance of the plane through (2, 3, - 1) and at right angles to the vector 3 hati - 4 hatj + 7 hatk from the origin is (A)  - (13/14) (B) (1/√74) (C) (13/√74) (D) none of these

Tardigrade Admin · Accepted Answer

The equation of the plane through (2, 3, - 1) , and perpendicular to the vector 3 hati - 4 hatj + 7 hatk is 3 (x - 2) + (- 4) (y - 3) + 7 (z - (- 1)) = 0 or 3 x - 4 y + 7 z + 13 = 0 Distance of this plane from the origin = (|3 ×0-4 ×0 + 7 × 0+13|/√32 +(-4)2 +72)

Q. The distance of the plane through $(2, 3, - 1)$ and at right angles to the vector $3 \hat{i} - 4 \hat{j} + 7 \hat{k}$ from the origin is

$- \frac{13}{14}$

$\frac{1}{74}$

$\frac{13}{74}$

none of these

Q. The distance of the plane through (2,3,−1) and at right angles to the vector 3i^−4j^​+7k^ from the origin is

−1413​

74​1​

74​13​

none of these

The equation of the plane through (2,3,−1) , and perpendicular to the vector 3i^−4j^​+7k^ is 3(x−2)+(−4)(y−3)+7(z−(−1))=0 or 3x−4y+7z+13=0 Distance of this plane from the origin =32+(−4)2+72​∣3×0−4×0+7×0+13∣​

Q. The distance of the plane through $(2, 3, - 1)$ and at right angles to the vector $3 \hat{i} - 4 \hat{j} + 7 \hat{k}$ from the origin is

$- \frac{13}{14}$

$\frac{1}{74}$

$\frac{13}{74}$