First slide

Dot product or scalar product of vectors

Question

$\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{c} = \hat{i} + \hat{j} - 2 \hat{k}$ . A vector co-planar with $\vec{b} and \vec{c}$ whose projection on $\vec{a}$ is magnitude $\sqrt{\frac{2}{3}}$ is

Moderate

A

$2 \hat{i} + 3 \hat{j} - 3 \hat{k}$
B

$- 2 \hat{i} - \hat{j} + 5 \hat{k}$
C

$2 \hat{i} + 3 \hat{j} + 3 \hat{k}$
D

$2 \hat{i} + \hat{j} + 5 \hat{k}$

Solution

Let the required vector be $\vec{r}$ then
$\begin{array}{l} \vec{r} = x_{1} \vec{b} + x_{2} \vec{c} and \vec{r} \cdot \vec{a} = \sqrt{\frac{2}{3}} (| \vec{a} |) = 2 \\ Now, \vec{r} \cdot \vec{a} = x_{1} \vec{a} \cdot \vec{b} + x_{2} \vec{a} \cdot \vec{c} \\ \Rightarrow 2 = x_{1} (2 - 2 - 1) + x_{2} (2 - 1 - 2) \\ \Rightarrow x_{1} + x_{2} = - 2 \\ \Rightarrow \vec{r} = x_{1} (\hat{i} + 2 \hat{j} - \hat{k}) + x_{2} (\hat{i} + \hat{j} - 2 \hat{k}) \end{array}$
$\begin{array}{l} = \hat{i} (x_{1} + x_{2}) + \hat{j} (2 x_{1} + x_{2}) - \hat{k} (2 x_{2} + x_{1}) \\ = - 2 \hat{i} + \hat{j} (x_{1} - 2) - \hat{k} (- 4 - x_{1}) \end{array}$

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