If a→=i^+2j^+3k^, b→=-i^+2j^+k^, c→=3i^+j^ and a→+pb→ is normal to c→, then p is equal to

Correct answer: (C) 5. a→+pb→ is normal to c→ ∴ a→+pb→·c→=0 1-pi^+2+2pj^+3+pk^·3i^+j^=0 3-3p+2+2p=0⇒p=5

Q79 / 125 · BITSAT 2019 · Mathematics

If $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k}, \vec{c} = 3 \hat{i} + \hat{j}$ and $\vec{a} + p \vec{b}$ is normal to $\vec{c}$ , then $p$ is equal to

$(\vec{a} + p \vec{b})$ is normal to $\vec{c}$

$∴ (\vec{a} + p \vec{b}) \cdot \vec{c} = 0$

$((1 - p) \hat{i} + (2 + 2 p) \hat{j} + (3 + p) \hat{k}) \cdot (3 \hat{i} + \hat{j}) = 0$

$3 - 3 p + 2 + 2 p = 0 \Rightarrow p = 5$

Hence, the correct answer is option: 'C'.