Vectors 2 Question 4
4. Let $\vec{\alpha}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\vec{\beta}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$. If $\vec{\beta}=\vec{\beta} _1-\vec{\beta} _2$, where $\vec{\beta} _1$ is parallel to $\vec{\alpha}$ and $\vec{\beta} _2$ is perpendicular to $\vec{\alpha}$, then $\vec{\beta} _1 \times \vec{\beta} _2$ is equal to
(a) $\frac{1}{2}(3 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})$
(b) $\frac{1}{2}(-3 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})$
(c) $-3 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
(d) $3 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$
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Answer:
Correct Answer: 4. (c)
Solution:
- Given vectors $\vec{\alpha}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\vec{\beta}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\vec{\beta}=\vec{\beta} _1-\vec{\beta} _2$ such that $\vec{\beta} _1$ is parallel to $\vec{\alpha}$ and $\vec{\beta} _2$ is perpendicular to $\alpha$
$ \begin{aligned} & \text { So, } \quad \vec{\beta} _1=\lambda \alpha=\lambda(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}) \\ & \text { Now, } \quad \vec{\beta} _2=\vec{\beta} _1-\vec{\beta}=\lambda(3 \hat{\mathbf{i}}+\hat{\mathbf{j}})-(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ & =(3 \lambda-2) \hat{\mathbf{i}}+(\lambda+1) \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $
$\because \vec{\beta} _2$ is perpendicular to $\alpha$, so $\vec{\beta} _2 \cdot \alpha=0$
[since if non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other, then $\mathbf{a} \cdot \mathbf{b}=0]$
$ \begin{array}{lr} \therefore & (3 \lambda-2)(3)+(\lambda+1)(1)=0 \\ \Rightarrow & 9 \lambda-6+\lambda+1=0 \\ \Rightarrow & 10 \lambda=5 \Rightarrow \lambda=\frac{1}{2} \end{array} $
So, $\vec{\beta} _1=\frac{3}{2} \hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}$
and $\vec{\beta} _2=\left(\frac{3}{2}-2\right) \hat{\mathbf{i}}+\left(\frac{1}{2}+1\right) \hat{\mathbf{j}}-3 \hat{\mathbf{k}}=-\frac{1}{2} \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
$ \begin{aligned} \therefore \vec{\beta} _1 \times \vec{\beta} _2 & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ \frac{-1}{2} & \frac{3}{2} & -3 \end{array}\right|=\hat{\mathbf{i}}\left(-\frac{3}{2}-0\right)-\hat{\mathbf{j}}\left(-\frac{9}{2}-0\right)+\hat{\mathbf{k}}\left(\frac{9}{4}+\frac{1}{4}\right) \\ & =-\frac{3}{2} \hat{\mathbf{i}}+\frac{9}{2} \hat{\mathbf{j}}+\frac{5}{2} \hat{\mathbf{k}}=\frac{1}{2}(-3 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \end{aligned} $
