Question

show. that. vectota=2i+3j+6k. b=3i-6j+2


k. c=6i+2j-3k are mutually. perpendicular
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Answer 1

Explanation:

Two vectors are perpendicular if there dot product is zero.

$\vec A.\vec B= 0 \\ \\ \vec A = 2 \hat i + 3 \hat j + 6\hat k \\ \\ \vec B = 3\hat i - 6 \hat j + 2\hat k \\ \\\vec A.\vec B = 2(3) + (3)( - 6) + (6)(2) \\ \\ = 6 - 18 + 12 \\ \\ \vec A.\vec B= 0$

Thus vector A and vector B are perpendicular to each other.

$\vec A.\vec C = 0 \\ \\ \vec A= 2 \hat i + 3 \hat j + 6\hat k \\ \\ \vec C = 6\hat i + 2\hat j - 3\hat k \\ \\\vec A.\vec C = 6(2) + (3)( 2) + (6)( - 3) \\ \\ = 12 + 6 - 18 \\ \\ \vec A.\vec C= 0$

Thus vector A and vector C are perpendicular to each other.

$\vec B.\vec C = 0 \\ \\ \vec B = 3\hat i - 6 \hat j + 2\hat k \\ \\ \vec C = 6\hat i + 2\hat j - 3\hat k \\ \\\vec B.\vec C = 6(3) + ( - 6)( 2) + (2)( - 3) \\ \\ = 18 - 12 - 6 \\ \\ \vec B.\vec C= 0$

Thus vector B and vector C are perpendicular to each other.

Thus all vectors are mutually perpendicular to each other.

Hope it helps you.