
The point of intersection of the lines
\[\begin{align}
& {{{l}}_{1}}=\,\overset{\to }{\mathop{{r}}}\,{(t)}\,{=}\,{(\hat{i}-6\hat{j}+2\hat{k}) + t(\hat{i}}+2{\hat{j}}+\,{\hat{k}}) \\
& {{\text{l}}_{2}}=\overset{\to }{\mathop{{R}}}\,{(u)}\,\text{=}\,{(4\hat{j}}+{\hat{k}})\,+\,{u(2\hat{i}}+{\hat{j}+2\hat{k}})\,\text{is} \\
\end{align}\]
A.(4,4,5)
B.(6,4,7)
C.(8,8,9)
D.(10,12,11)
Answer
509.1k+ views
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:
\[\begin{align}
& {{{l}}_{1}}=\,{(\hat{i}-6\hat{j}+2\hat{k}) + t(\hat{i}}+2{\hat{j}}+\,{\hat{k}}) \\
& {{{l}}_{1}}=\,{(1+t)\hat{i} + t(2t -}\,{6}){\hat{j} + (2+t)\hat{k}}(1) \\
& {{{l}}_{2}}=\,{(4\hat{j}}+{\hat{k}})\,+\,\mu {(2\hat{i}}+{\hat{j}+2\hat{k}}) \\
& {{{l}}_{2}}=\,{(2}\mu )\,+\,(4+\mu ){\hat{j}}\,+\,{(1}+{2}\mu ){\hat{k}}(2) \\
\end{align}\]
At the point of intersection, coordinates of l$_{1}$ must be respectively equal to that of l$_{2}$
Comparing (1) & (2)
\[\begin{align}
& \text{we}\,\text{get (l+t)}\,\text{=}\,\text{(2}\mu \text{)}\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(3)} \\
& \text{(2t-6)}\,\text{=}\,\text{(4+}\mu \text{)}\text{(4)} \\
& \text{(2+t) = (1+2}\mu \text{)}\text{(5)}
\end{align}\]
Solving (3) & (4) gives
\[\begin{align}
& 3\text{t}\,\text{=}\,\text{2l}\,\Rightarrow \,\text{t}\,\text{=}\,\text{7} \\
& \text{2}\mu \,=\,8\,\Rightarrow \,\mu \,=\,4
\end{align}\]
$\therefore \,(\text{t,}\,\mu \text{)}\,\text{=}\,\text{(7,}\,\text{4)}$ satisfied (5) there is a unique point of intersection substituting back in eq (1) or eq (2)
\[\begin{align}
& \text{(x,}\,\text{y,}\,\text{z)}\,\text{=}\,\text{(1+7,}\,\text{14 - 6,}\,\text{2+7)} \\
& \text{(x,}\,\text{y,}\,\text{z)}\,\text{=}\,\text{(8,}\,\text{8,}\,\text{9)} \\
\end{align}\]
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.
Complete step-by-step answer:
\[\begin{align}
& {{{l}}_{1}}=\,{(\hat{i}-6\hat{j}+2\hat{k}) + t(\hat{i}}+2{\hat{j}}+\,{\hat{k}}) \\
& {{{l}}_{1}}=\,{(1+t)\hat{i} + t(2t -}\,{6}){\hat{j} + (2+t)\hat{k}}(1) \\
& {{{l}}_{2}}=\,{(4\hat{j}}+{\hat{k}})\,+\,\mu {(2\hat{i}}+{\hat{j}+2\hat{k}}) \\
& {{{l}}_{2}}=\,{(2}\mu )\,+\,(4+\mu ){\hat{j}}\,+\,{(1}+{2}\mu ){\hat{k}}(2) \\
\end{align}\]
At the point of intersection, coordinates of l$_{1}$ must be respectively equal to that of l$_{2}$
Comparing (1) & (2)
\[\begin{align}
& \text{we}\,\text{get (l+t)}\,\text{=}\,\text{(2}\mu \text{)}\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(3)} \\
& \text{(2t-6)}\,\text{=}\,\text{(4+}\mu \text{)}\text{(4)} \\
& \text{(2+t) = (1+2}\mu \text{)}\text{(5)}
\end{align}\]
Solving (3) & (4) gives
\[\begin{align}
& 3\text{t}\,\text{=}\,\text{2l}\,\Rightarrow \,\text{t}\,\text{=}\,\text{7} \\
& \text{2}\mu \,=\,8\,\Rightarrow \,\mu \,=\,4
\end{align}\]
$\therefore \,(\text{t,}\,\mu \text{)}\,\text{=}\,\text{(7,}\,\text{4)}$ satisfied (5) there is a unique point of intersection substituting back in eq (1) or eq (2)
\[\begin{align}
& \text{(x,}\,\text{y,}\,\text{z)}\,\text{=}\,\text{(1+7,}\,\text{14 - 6,}\,\text{2+7)} \\
& \text{(x,}\,\text{y,}\,\text{z)}\,\text{=}\,\text{(8,}\,\text{8,}\,\text{9)} \\
\end{align}\]
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.
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