Question

The point of intersection of the lines

& {{{l}}_{1}}=\,\overset{\to }{\mathop{{r}}}\,{(t)}\,{=}\,{(\hat{i}-6\hat{j}+2\hat{k}) + t(\hat{i}}+2{\hat{j}}+\,{\hat{k}}) \\\ & {{\text{l}}_{2}}=\overset{\to }{\mathop{{R}}}\,{(u)}\,\text{=}\,{(4\hat{j}}+{\hat{k}})\,+\,{u(2\hat{i}}+{\hat{j}+2\hat{k}})\,\text{is} \\\ \end{aligned}$$ A.(4,4,5) B.(6,4,7) C.(8,8,9) D.(10,12,11)

Answer 1

Hint: At the point of intersection there is a common point for both the lines. Hence they should satisfy both the equations.

Complete step-by-step answer:

& {{{l}}_{1}}=\,{(\hat{i}-6\hat{j}+2\hat{k}) + t(\hat{i}}+2{\hat{j}}+\,{\hat{k}}) \\\ & {{{l}}_{1}}=\,{(1+t)\hat{i} + t(2t -}\,{6}){\hat{j} + (2+t)\hat{k}}(1) \\\ & {{{l}}_{2}}=\,{(4\hat{j}}+{\hat{k}})\,+\,\mu {(2\hat{i}}+{\hat{j}+2\hat{k}}) \\\ & {{{l}}_{2}}=\,{(2}\mu )\,+\,(4+\mu ){\hat{j}}\,+\,{(1}+{2}\mu ){\hat{k}}(2) \\\ \end{aligned}$$ At the point of intersection, coordinates of l$_{1}$ must be respectively equal to that of l$_{2}$ Comparing (1) & (2) $$\begin{aligned} & \text{we}\,\text{get (l+t)}\,\text{=}\,\text{(2}\mu \text{)}\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(3)} \\\ & \text{(2t-6)}\,\text{=}\,\text{(4+}\mu \text{)}\text{(4)} \\\ & \text{(2+t) = (1+2}\mu \text{)}\text{(5)} \end{aligned}$$ Solving (3) & (4) gives $$\begin{aligned} & 3\text{t}\,\text{=}\,\text{2l}\,\Rightarrow \,\text{t}\,\text{=}\,\text{7} \\\ & \text{2}\mu \,=\,8\,\Rightarrow \,\mu \,=\,4 \end{aligned}$$ $\therefore \,(\text{t,}\,\mu \text{)}\,\text{=}\,\text{(7,}\,\text{4)}$ satisfied (5) there is a unique point of intersection substituting back in eq (1) or eq (2) $$\begin{aligned} & \text{(x,}\,\text{y,}\,\text{z)}\,\text{=}\,\text{(1+7,}\,\text{14 - 6,}\,\text{2+7)} \\\ & \text{(x,}\,\text{y,}\,\text{z)}\,\text{=}\,\text{(8,}\,\text{8,}\,\text{9)} \\\ \end{aligned}$$ Note: While solving for the parameters (t, $\mu $) we have 3 equations for 2 variables if the values of (t, $\mu $) that are obtained from (1) & (2) doesn’t satisfy the (3)rd eqn then there is no point of intersection. In the above case we get only a single point of intersection.