A unit vector perpendicular to the plane formed by the points (1,0,1), (0, 2,2) and (3,3, 0) is

Question

A unit vector perpendicular to the plane formed by the points (1,0,1), (0, 2,2) and (3,3, 0) is (A) (1/5√3)(5 hati- hatj-7 hatk) (B) (1/5√3)(5 hati- hatj+7 hatk) (C) (1/5√3)(5 hati+ hatj+7 hatk) (D) None of these

Tardigrade Admin · Accepted Answer

Let A(1, 0, 1), B(0,2,2) and C (3,3, 0) be the given points, then AB = - hati+2 hatj+ hatk, BC = 3 hati+ hatj-2 hatk AB× BC = | hati& hatj& hatk -1&2&1 3&1&-2| = -5 hati+ hatj-7 hatk ∴ unit vector ⊥ to the plane: ABC = ±(1/5√3)(-5 hati+ hatj-7 hatk)

Q. A unit vector perpendicular to the plane formed by the points $(1, 0, 1), (0, 2, 2)$ and $(3, 3, 0)$ is

$\frac{1}{5 3} (5 \hat{i} - \hat{j} - 7 \hat{k})$

$\frac{1}{5 3} (5 \hat{i} - \hat{j} + 7 \hat{k})$

$\frac{1}{5 3} (5 \hat{i} + \hat{j} + 7 \hat{k})$

None of these

Q. A unit vector perpendicular to the plane formed by the points (1,0,1),(0,2,2) and (3,3,0) is

53​1​(5i^−j^​−7k^)

53​1​(5i^−j^​+7k^)

53​1​(5i^+j^​+7k^)

None of these

Let A(1,0,1),B(0,2,2) and C(3,3,0) be the given points, then AB=−i^+2j^​+k^,BC=3i^+j^​−2k^ AB×BC=∣∣​i^−13​j^​21​k^1−2​∣∣​=−5i^+j^​−7k^ ∴ unit vector ⊥ to the plane: ABC =±53​1​(−5i^+j^​−7k^)

Q. A unit vector perpendicular to the plane formed by the points $(1, 0, 1), (0, 2, 2)$ and $(3, 3, 0)$ is

$\frac{1}{5 3} (5 \hat{i} - \hat{j} - 7 \hat{k})$

$\frac{1}{5 3} (5 \hat{i} - \hat{j} + 7 \hat{k})$

$\frac{1}{5 3} (5 \hat{i} + \hat{j} + 7 \hat{k})$