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If $\alpha=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}, \beta=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$, then express $\beta$ in the form $\beta=\beta_1+\beta_2$ where $\beta_1$ is parallel to $\alpha$ and $\beta_2$ is perpendicular to $\alpha$, then $\beta_1$ is given by
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Given $\alpha=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}, \boldsymbol{\beta}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$$
\beta=\beta_1+\beta_2
$$
$\beta_1$ is parallel to $\alpha$
$$
\begin{aligned}
\Rightarrow \quad \beta_1 & =\lambda \alpha \Rightarrow \beta_1=\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}} \\
\Rightarrow \text { Also, } \beta & =\beta_1+\beta_2 \\
\beta_2 & =\beta-\boldsymbol{\beta}_1=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})-(\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}}) \\
\boldsymbol{\beta}_2 & =(1-\lambda) \hat{\mathbf{i}}+(2-3 \lambda) \hat{\mathbf{j}}-\hat{\mathbf{k}}
\end{aligned}
$$
It is given $\beta_2$ is perpendicular to $\alpha$.
$$
\begin{array}{cc}
\therefore(1-\lambda)(1)+(2+3 \lambda)(-3)+(-1)(0)=0 \\
\Rightarrow & 1-\lambda-6-9 \lambda+0=0 \\
\Rightarrow & -5-10 \lambda=0 \\
\Rightarrow & \lambda=-\frac{1}{2} \\
\beta_1=\frac{-1}{2} \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}
\end{array}
$$
$$
\beta=\beta_1+\beta_2
$$
$\beta_1$ is parallel to $\alpha$
$$
\begin{aligned}
\Rightarrow \quad \beta_1 & =\lambda \alpha \Rightarrow \beta_1=\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}} \\
\Rightarrow \text { Also, } \beta & =\beta_1+\beta_2 \\
\beta_2 & =\beta-\boldsymbol{\beta}_1=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})-(\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}}) \\
\boldsymbol{\beta}_2 & =(1-\lambda) \hat{\mathbf{i}}+(2-3 \lambda) \hat{\mathbf{j}}-\hat{\mathbf{k}}
\end{aligned}
$$
It is given $\beta_2$ is perpendicular to $\alpha$.
$$
\begin{array}{cc}
\therefore(1-\lambda)(1)+(2+3 \lambda)(-3)+(-1)(0)=0 \\
\Rightarrow & 1-\lambda-6-9 \lambda+0=0 \\
\Rightarrow & -5-10 \lambda=0 \\
\Rightarrow & \lambda=-\frac{1}{2} \\
\beta_1=\frac{-1}{2} \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}
\end{array}
$$
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