If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+t \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2…
If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+t \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$, then the values of ' $t$ ' for which (a+b) and (a-b) are perpendicular, are
$\pm 2$
$\pm 2 \sqrt{3}$
$\pm 3 \sqrt{2}$
$\pm 3$
Solution
We have,
$\begin{aligned} \mathbf{a}+\mathbf{b} & =(\mathbf{i}+\mathbf{j}+t \mathbf{k})+(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}) \\ & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+(t+3) \hat{\mathbf{k}}\end{aligned}$
and
$\begin{aligned} \mathbf{a}-\mathbf{b} & =(\hat{\mathbf{i}}+\hat{\mathbf{j}}+t \hat{\mathbf{k}})-(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ & =-\hat{\mathbf{j}}+(t-3) \hat{\mathbf{k}}\end{aligned}$
Since, $(\mathbf{a}+\mathbf{b})$ and $(\mathbf{a}-\mathbf{b})$ are perpendicular to each other, then
$(\mathbf{a}+\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})=0$
$\begin{array}{cc}\Rightarrow & {[2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+(t+3) \hat{\mathbf{k}}] \cdot[-\hat{\mathbf{j}}+(t-3) \hat{\mathbf{k}}]=0} \\ \Rightarrow & 0-3+t^2-9=0 \\ \Rightarrow & \quad 0\end{array}$
$\Rightarrow \quad t^2=12 \Rightarrow t= \pm 2 \sqrt{3}$