Vectors 1 Question 4
4. Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\sqrt{2} \hat{\mathbf{k}}, \mathbf{b}=b _1 \hat{\mathbf{i}}+b _2 \hat{\mathbf{j}}+\sqrt{2} \hat{\mathbf{k}} \quad$ and $\mathbf{c}=5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\sqrt{2} \hat{\mathbf{k}}$ be three vectors such that the projection vector of $\mathbf{b}$ on $\mathbf{a}$ is $\mathbf{a}$. If $\mathbf{a}+\mathbf{b}$ is perpendicular to $\mathbf{c}$, then $|\mathbf{b}|$ is equal to
(a) 6
(b) 4
(c) $\sqrt{22}$
(d) $\sqrt{32}$
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Answer:
Correct Answer: 4. (b)
Solution:
- According to given information, we have the following figure.
Clearly, projection of $b$ on $a=\frac{b \cdot a}{|a|}$
$$ \begin{aligned} & =\frac{\left(b _1 \hat{i}+b _2 \hat{j}+\sqrt{2} \hat{k}\right)(\hat{i}+\hat{j}+\sqrt{2} \hat{k})}{\sqrt{1^{2}+1^{2}+(\sqrt{2})^{2}}} \\ & =\frac{b _1+b _2+2}{\sqrt{4}}=\frac{b _1+b _2+2}{2} \end{aligned} $$
But projection of $b$ on $a=|a|$
$\therefore \frac{b _1+b _2+2}{2}=\sqrt{1^{2}+1^{2}+(\sqrt{2})^{2}}$
$\Rightarrow \frac{b _1+b _2+2}{2}=2 \Rightarrow b _1+b _2=2$
Now, $a+b=(\hat{i}+\hat{j}+\sqrt{2} \hat{k})+\left(b _1 \hat{i}+b _2 \hat{j}+\sqrt{2} \hat{k}\right)$
$$ =\left(b _1+1\right) \hat{i}+\left(b _2+1\right) \hat{j}+2 \sqrt{2} \hat{k} $$
$\because(a+b) \perp c$, therefore $(a+b) \cdot c=0$
$\Rightarrow {(b _1+1) \hat{i}+(b _2+1) \hat{j}+2 \sqrt{2} \hat{k}}(5 \hat{i}+\hat{j}+\sqrt{2} \hat{k})=0$
$\Rightarrow 5\left(b _1+1\right)+1\left(b _2+1\right)+2 \sqrt{2}(\sqrt{2})=0$
$\Rightarrow \quad 5 b _1+b _2=-10$
From Eqs. (i) and (ii), $b _1=-3$ and $b _2=5$
$\Rightarrow b=-3 \hat{i}+5 \hat{j}+\sqrt{2} \hat{k}$
$\Rightarrow|b|=\sqrt{(-3)^{2}+(5)^{2}+(\sqrt{2})^{2}}=\sqrt{36}=6$
5 Let $1^{\text {st }}$ line is $x=a y+b, z=c y+d$.
$\Rightarrow \frac{x-b}{a}=y, \frac{z-d}{c}=y \Rightarrow \frac{x-b}{a}=y=\frac{z-d}{c}$
The direction vector of this line is $b _1=a \hat{i}+\hat{j}+c \hat{k}$.
Let $2^{\text {nd }}$ line is $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$.
$\Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=z, \frac{y-d^{\prime}}{c^{\prime}}=z \Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=\frac{y-d^{\prime}}{c^{\prime}}=z$
The direction vector of this line is $b _2=a^{\prime} \hat{i}+c^{\prime} \hat{j}+\hat{k}$.
$\because$ The two lines are perpendicular, therefore, $b _1 \cdot b _2=0$.
$\Rightarrow(a \hat{i}+\hat{j}+c \hat{k}) \cdot\left(a^{\prime} \hat{i}+c^{\prime} \hat{j}+\hat{k}\right)=0$
$\Rightarrow a a^{\prime}+c^{\prime}+c=0 \quad \Rightarrow a a^{\prime}+c+c^{\prime}=0$
