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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 10 of 78
Marks: +1, -0
Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i^−j^+4k^\hat{2i} - \hat{j} + \hat{4k} and is in the direction i^+2j^−k^.\hat{i} + \hat{2j} - \hat{k} .
Solution:  
We have, a⃗=2i^−j^+4k^\vec{a} = \hat{2i} - \hat{j} + \hat{4k} , b⃗=i^+2j^−k^\vec{b} = \hat{i} + \hat{2j} - \hat{k}
Vector equation of the line is r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}
⇒r⃗=(2i^−j^+4k^)+λ(i^+2j^−k^)\Rightarrow \vec{r} = (\hat{2i} - \hat{j} + \hat{4k}) + \lambda (\hat{i} + \hat{2j} - \hat{k})
Now r⃗\vec{r} is the position vector of any point P(x, y, z) on the line.
∴xi^+yj^+zk^=(2i^−j^+4k^)+λ(i^+2j^−k^)\therefore \hat{xi} + \hat{yj} + \hat{zk} = (\hat{2i} - \hat{j} + \hat{4k}) + \lambda (\hat{i} + \hat{2j} - \hat{k})
⇒xi^+yj^+zk^=(2+λ)i^+(−1+2λ)j^+(4−λ)k^)\Rightarrow \hat{xi} + \hat{yj} + \hat{zk} = (2+\lambda)\hat{i} + (-1+2\lambda)\hat{j} + (4-\lambda)\hat{k} )
Eliminating λ, we get x−21=y+12=−z−41\frac{x-2}{1} = \frac{y+1}{2} = -\frac{z-4}{1} is the cartesian equation of the line.
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