JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 4)
Let $$\overrightarrow \alpha = 4\widehat i + 3\widehat j + 5\widehat k$$ and $$\overrightarrow \beta = \widehat i + 2\widehat j - 4\widehat k$$. Let $${\overrightarrow \beta _1}$$ be parallel to $$\overrightarrow \alpha $$ and $${\overrightarrow \beta _2}$$ be perpendicular to $$\overrightarrow \alpha $$. If $$\overrightarrow \beta = {\overrightarrow \beta _1} + {\overrightarrow \beta _2}$$, then the value of $$5{\overrightarrow \beta _2}\,.\left( {\widehat i + \widehat j + \widehat k} \right)$$ is :
9
7
6
11
Explanation
Let $\vec{\beta}_1=\lambda \vec{\alpha}$
Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$
$$ \begin{aligned} & =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})-\lambda(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \\\\ & =(1-4 \lambda) \hat{\mathrm{i}}+(2-3 \lambda) \hat{\mathrm{j}}-(5 \lambda+4) \hat{\mathrm{k}} \\\\ & \vec{\beta}_2 \cdot \vec{\alpha}=0 \\\\ & \Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0 \\\\ & \Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0 \\\\ & \Rightarrow 50 \lambda=-10 \\\\ & \Rightarrow \lambda=\frac{-1}{5} \\\\ & \vec{\beta}_2=\left(1+\frac{4}{5}\right) \hat{\mathrm{i}}+\left(2+\frac{3}{5}\right) \hat{\mathrm{j}}-(-1+4) \hat{\mathrm{k}} \\\\ & \vec{\beta}_2=\frac{9}{5} \hat{\mathrm{i}}+\frac{13}{5} \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\\\ & 5 \vec{\beta}_2=9 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\\\ & 5 \vec{\beta}_2 \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=9+13-15=7 \end{aligned} $$
Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$
$$ \begin{aligned} & =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})-\lambda(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \\\\ & =(1-4 \lambda) \hat{\mathrm{i}}+(2-3 \lambda) \hat{\mathrm{j}}-(5 \lambda+4) \hat{\mathrm{k}} \\\\ & \vec{\beta}_2 \cdot \vec{\alpha}=0 \\\\ & \Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0 \\\\ & \Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0 \\\\ & \Rightarrow 50 \lambda=-10 \\\\ & \Rightarrow \lambda=\frac{-1}{5} \\\\ & \vec{\beta}_2=\left(1+\frac{4}{5}\right) \hat{\mathrm{i}}+\left(2+\frac{3}{5}\right) \hat{\mathrm{j}}-(-1+4) \hat{\mathrm{k}} \\\\ & \vec{\beta}_2=\frac{9}{5} \hat{\mathrm{i}}+\frac{13}{5} \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\\\ & 5 \vec{\beta}_2=9 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\\\ & 5 \vec{\beta}_2 \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=9+13-15=7 \end{aligned} $$
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