Physics - Vectors Question with Solution | TestHub

Three vectors are mutually perpendicular, so the scalar or dot product of two vectors will zero. So, $\mathbf{A}\cdot \mathbf{B}=0,\mathbf{BC}=0,$ So, $\mathbf{A}\cdot \mathbf{B}=(a\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})\cdot (\hat{\mathbf{i}}+b\hat{\mathbf{j}}+\hat{\mathbf{k}})=0$ $\Rightarrow a+b+1=0$ $\mathbf{B}\cdot \mathbf{C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\cdot \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}=$ $\Rightarrow 1+b+c=0$ $\mathbf{A}\cdot \mathbf{C}=(a\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})\cdot (\hat{\mathbf{i}}+\hat{\mathbf{j}}+c\hat{\mathbf{k}})=0$ $\mathbf{A}\cdot \mathbf{C}=0$ $\Rightarrow a+1+c=0$ (iii Addition of Eqs. (i), (ii) and (iii), $\begin{array}{l}2(a+b+c)+3=0\\ a+b+c=-\frac{3}{2}\end{array}$ $\Rightarrow -1+c=-\frac{3}{2}$ $\begin{array}{rl}c& =-\frac{3}{2}+1\\ & =-\frac{1}{2}=0.5\\ a& =-\frac{1}{2}\\ b& =-\frac{1}{2}\\ c& =-\frac{1}{2}\end{array}$