Q73 / 120 · VITEEE 2010 · Mathematics

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

  • A
    2
  • B
    4
  • C
    6
  • Correct answer
    D
    8

Solution

$$ \begin{array}{l} \vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k} \\ \vec{b}=-\hat{i}+2 \hat{j}+\hat{k} \\ \vec{c}=3 \hat{i}+\hat{j} \end{array} $$ Now, $$ \begin{aligned} \vec{a}+t \vec{b} &=(2 \hat{i}+2 \hat{j}+3 \hat{k})+(-t \hat{i}+2 \hat{j}+t \hat{k}) \\ &=(2-t) \hat{i}+(2+2 t) \hat{j}+(3+t) \hat{k} \end{aligned} $$ Since, $\vec{a}+t \vec{b}$ is perpendicular to $\vec{c}$ $$ \begin{array}{l} (\vec{a}+t \vec{b}) \cdot \vec{c}=0 \\ \Rightarrow\{(2-t) \hat{i}+(2+2 t) \hat{j}+(3+t) \hat{k}\} \cdot(3 \hat{i}+\hat{j})=0 \\ \Rightarrow 3(2-t)+2+2 t=0 \\ \Rightarrow 6-3 t+2+2 t=0 \Rightarrow t=8 \end{array} $$

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

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