JEE MAIN - Mathematics (2003 - No. 26)
Let $$\overrightarrow u = \widehat i + \widehat j,\,\overrightarrow v = \widehat i - \widehat j$$ and $$\overrightarrow w = \widehat i + 2\widehat j + 3\widehat k\,\,.$$ If $$\widehat n$$ is a unit vector such that $$\overrightarrow u .\widehat n = 0$$ and $$\overrightarrow v .\widehat n = 0\,\,,$$ then $$\left| {\overrightarrow w .\widehat n} \right|$$ is equal to :
$$3$$
$$0$$
$$1$$
$$2$$
Explanation
Since, $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{u}}$ and $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{v}}$
$$ \begin{aligned} \Rightarrow \hat{\mathbf{n}} & =\frac{\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \\\\ \therefore|\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}| & =\left|\frac{\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|}\right| \\\\ & =\frac{|\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \end{aligned} $$
$$ \begin{aligned} & \text { Now, } \overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})=\left|\begin{array}{rrr} 1 & 2 & 3 \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{array}\right|=-6 \\\\ & \begin{array}{ll} \Rightarrow |\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|=6 \\\\ \text { and } \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-2 \hat{\mathbf{k}} \\\\ \Rightarrow |\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|=2 \\\\ \therefore |\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}|=\frac{6}{2}=3 \end{array} \end{aligned} $$
$$ \begin{aligned} \Rightarrow \hat{\mathbf{n}} & =\frac{\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \\\\ \therefore|\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}| & =\left|\frac{\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|}\right| \\\\ & =\frac{|\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \end{aligned} $$
$$ \begin{aligned} & \text { Now, } \overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})=\left|\begin{array}{rrr} 1 & 2 & 3 \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{array}\right|=-6 \\\\ & \begin{array}{ll} \Rightarrow |\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|=6 \\\\ \text { and } \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-2 \hat{\mathbf{k}} \\\\ \Rightarrow |\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|=2 \\\\ \therefore |\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}|=\frac{6}{2}=3 \end{array} \end{aligned} $$
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