Find the value of m for which the vectors a = 2i + mj - 3k and b = i - 2j + k are perpendicular.

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Solution

For two vectors to be perpendicular-

$a . b = 0$

$⟹ (2 i + m j - 3 k) . (i - 2 j + k) = 0$

$⟹ 2 - 2 m - 3 = 0$

$⟹ m = \frac{- 1}{2}$

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\vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}

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b = 2 i - 2 j + k

c = 2 i + 2 j + 3 k

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\to a = 2 ¯ i - 2 ¯ j + ¯ ¯ ¯ k

¯ ¯ b = 2 ¯ i + 2 ¯ j + 3 ¯ ¯ ¯ k

\vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}

\vec{a} = \hat{i} - \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - \hat{k}, \vec{c} = λ \hat{i} - \hat{j} + λ \hat{k}

\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{c} = λ \hat{i} + λ \hat{j} + 5 \hat{k}

\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{b} = 3 \hat{i} + λ \hat{j} + \hat{k}, \vec{c} = \hat{i} + 2 \hat{j} + 2 \hat{k}

\vec{a} = \hat{i} + 3 \hat{j}, \vec{b} = 5 \hat{k}, \vec{c} = λ \hat{i} - \hat{j}

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