JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 12)

Let $$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$$ and let $$\vec{b}$$ be a vector such that $$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$$ and $$\vec{a} \cdot \vec{b}=3$$. Then the projection of $$\vec{b}$$ on the vector $$\vec{a}-\vec{b}$$ is :
$$\frac{2}{\sqrt{21}}$$
$$2 \sqrt{\frac{3}{7}}$$
$$ \frac{2}{3} \sqrt{\frac{7}{3}} $$
$$\frac{2}{3}$$

Explanation

$$\overrightarrow a = \widehat i - \widehat j + 2\widehat k$$

$$\overrightarrow a \times \overrightarrow b = 2\widehat i - \widehat k$$

$$\overrightarrow a \,.\,\overrightarrow b = 3$$

$$|\overrightarrow a \times \overrightarrow b {|^2} + |\overrightarrow a \,.\,\overrightarrow b {|^2} = |\overrightarrow a {|^2}\,.\,|\overrightarrow b {|^2}$$

$$ \Rightarrow 5 + 9 = 6|\overrightarrow b {|^2}$$

$$ \Rightarrow |b {|^2} = {7 \over 3}$$

$$|\overrightarrow a - \overrightarrow b | = \sqrt {|\overrightarrow a {|^2} + |\overrightarrow b {|^2} - 2\overrightarrow a \,.\,\overrightarrow b } = \sqrt {{7 \over 3}} $$

projection of $$\overrightarrow b $$ on $$\overrightarrow a - \overrightarrow b = {{\overrightarrow b \,.\,(\overrightarrow a - \overrightarrow b )} \over {|\overrightarrow a - \overrightarrow b |}}$$

$$ = {{\overrightarrow b \,.\,\overrightarrow a - |\overrightarrow b {|^2}} \over {|\overrightarrow a - \overrightarrow b |}} = {{3 - {7 \over 3}} \over {\sqrt {{7 \over 3}} }}$$

$$ = {2 \over {\sqrt {21} }}$$

Comments (0)

Advertisement