If a = hati + hatj + hatk and b = hatj - hatk, then find a vector c s…

The correct answer is Option (2) → $\frac{1}{3}(5\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}})$ ##

Let $\mathbf{c} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}$

Also, $\mathbf{a} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$ and $\mathbf{b} = \hat{\mathbf{j}} - \hat{\mathbf{k}}$

For $\mathbf{a} \times \mathbf{c} = \mathbf{b}$,

$\begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = \hat{\mathbf{j}} - \hat{\mathbf{k}}$

$\Rightarrow \hat{\mathbf{i}}(z - y) - \hat{\mathbf{j}}(z - x) + \hat{\mathbf{k}}(y - x) = \hat{\mathbf{j}} - \hat{\mathbf{k}}$

On comparing both sides, we get

$∴z - y = 0 \quad \dots(i)$

$x - z = 1 \quad \dots(ii)$

$y - x = -1 \quad \dots(iii)$

Also, $\mathbf{a} \cdot \mathbf{c} = 3$ [given]

$(\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) \cdot (x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}) = 3 \quad \dots(iv)$

$\Rightarrow x + y + z = 3$

On adding Eqs. $(ii)$ and $(iii)$, we get

$2x - y - z = 2 \quad \dots(v)$

On adding Eqs. $(iv)$ and $(v)$, we get

$3x = 5$

$x = \frac{5}{3}$

On putting the value of $x$ in Eq. $(ii)$ to evaluate the value of $y, z$, we get

$∴y = \frac{5}{3} - 1 = \frac{2}{3} \text{ and } z = \frac{2}{3}$

Now, $\vec{c} = \frac{5}{3}\hat{\mathbf{i}} + \frac{2}{3}\hat{\mathbf{j}} + \frac{2}{3}\hat{\mathbf{k}}$

$= \frac{1}{3}(5\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}})$