Mathematics - Vector Question with Solution | TestHub

Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2\hat{\mathbf{k}},\mathbf{b}=\hat{\mathbf{i}}+2\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ $\therefore$ A vector coplanar to $\mathbf{a}$ and $\mathbf{b}$, and perpendicular to $\mathbf{c}$. Now, $\lambda (\mathbf{a}\times \mathbf{b})\times \mathbf{c}$ $\Rightarrow \lambda \{(\mathbf{a}\cdot \mathbf{c})\mathbf{b}-(\mathbf{b}\cdot \mathbf{c})\mathbf{a}\}$ $\Rightarrow \lambda \{(1+1+4)(\hat{\mathbf{i}}+2\hat{\mathbf{j}}+\hat{\mathbf{k}})$ $-(1+2+1)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2\hat{\mathbf{k}})\}$ $\Rightarrow \lambda \{6\hat{\mathbf{i}}+12\hat{\mathbf{j}}+6\hat{\mathbf{k}}-6\hat{\mathbf{i}}-6\hat{\mathbf{j}}-12\hat{\mathbf{k}}\}$ $\Rightarrow \lambda \{6\hat{\mathbf{j}}-6\hat{\mathbf{k}}\}\Rightarrow 6\lambda (\hat{\mathbf{j}}-\hat{\mathbf{k}})$ For $\lambda =\frac{1}{6}\Rightarrow$ Option (a) is correct. For $\lambda =-\frac{1}{6}\Rightarrow$ Option (d) is correct.