Vectors 5 Question 8
8. If $\overrightarrow{\mathbf{A}}=(1,1,1), \overrightarrow{\mathbf{C}}=(0,1,-1)$ are given vectors, then a vector $\overrightarrow{\mathbf{B}}$ satisfying the equations $\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{C}}$ and $\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}=3$ is ….
(1985,91, 2M)
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Answer:
Correct Answer: 8. $\left(\frac{5}{3} \hat{\mathbf{i}}, \frac{2}{3} \hat{\mathbf{j}}, \frac{2}{3} \hat{\mathbf{k}}\right)$
Solution:
- Let
$\overrightarrow{\mathbf{B}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
Given, $\quad \overrightarrow{\mathbf{A}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{C}}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$
Also, given $\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{C}}$
$\Rightarrow \quad(z-y) \hat{\mathbf{i}}-(z-x) \hat{\mathbf{j}}+(y-x) \hat{\mathbf{k}}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\Rightarrow \quad z-y=0, x-z=1, y-x=-1$
Also, $\quad \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}=3 \Rightarrow x+y+z=3$
On solving above equations, we get
$ \begin{aligned} & x & =\frac{5}{3}, y=z=\frac{2}{3} \\ \therefore \quad & \overrightarrow{\mathbf{B}} & =\left(\frac{5}{3} \hat{\mathbf{i}}, \frac{2}{3} \hat{\mathbf{j}}, \frac{2}{3} \hat{\mathbf{k}}\right) \end{aligned} $
