Mathematics - Vector Question with Solution | TestHub

Let the unit vector along $\overline{\mathrm{c}}$ be $\mathrm{x}\overline{\mathrm{i}}+\mathrm{y}\overline{\mathrm{j}}+\mathrm{z}\overline{\mathrm{k}}$ where ${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=1$

$$\begin{array}{rlr}& \therefore -\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}=\mathrm{t}\left[\mathrm{xi}+\mathrm{y}\overline{\mathrm{j}}+\mathrm{z}\overline{\mathrm{k}}+\frac{3\mathrm{i}+4\mathrm{j}}{5}\right]& \begin{array}{rlrlrlrl}\Rightarrow -\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}=\mathrm{t}& \left[\left(\frac{5\mathrm{x}+3}{5}\right)\overline{\mathrm{i}}+\left(\frac{5\mathrm{y}+4}{5}\right)\overline{\mathrm{j}}+\mathrm{z}\overline{\mathrm{k}}\right]\Rightarrow \left(\frac{5\mathrm{x}+3}{5}\right)\mathrm{t}& =-1,\left(\frac{5\mathrm{y}+4}{5}\right)\mathrm{t}=1,\mathrm{zt}=-1& {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=1& \Rightarrow {\left(-\frac{1}{\mathrm{t}}-\frac{3}{5}\right)}^2+{\left(\frac{1}{\mathrm{t}}-\frac{4}{5}\right)}^2+{\left(-\frac{1}{\mathrm{t}}\right)}^2=1& \Rightarrow \frac{3}{{\mathrm{t}}^2}+\frac{2}{\mathrm{t}}\left(\frac{3}{5}-\frac{4}{5}\right)+1=1& \Rightarrow \frac{3}{{\mathrm{t}}^2}-\frac{2}{5\mathrm{t}}=0\Rightarrow \overline{\mathrm{c}}=\frac{-1}{15}(11\overline{\mathrm{i}}+10\overline{\mathrm{j}}+2\overline{\mathrm{k}})\Rightarrow 15-2\mathrm{t}=0& \therefore \mathrm{t}=\frac{15}{2}\end{array}\end{array}$$