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Home » Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
CBSE | Class 12 | Exercise 27A | Maths | RS Aggarwal | Straight Line in Space
Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).

Find the coordinates of the foot of the perpendicular drawn from the point A(1,8,4) to the line joining the points B(0,-1,3) and C(2,-3,-1)
Answer
Given: perpendicular drawn from point A(1,8,4) to line joining points B(0,-1,3) and C(2,-3,-1)
To find: foot of perpendicular
Formula Used: Equation of a line is
Vector form: \vec{I}=\vec{a}+\lambda \vec{b}
Cartesian form: \frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{y-y_{1}}{\mathrm{~b}_{\mathrm{z}}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{~h}_{\mathrm{J}}}=\lambda
where \vec{a}=x_{1} \hat{\imath}+y_{1} \hat{l}+z_{1} \hat{k} is a point on the line and \vec{b}=b_{1} \hat{l}+b_{2} \hat{\jmath}+b_{3} \hat{k} with b_{1}: b_{2}: b_{3} being the direction ratios of the line.
If 2 lines of direction ratios a_{1}: a_{2}: a_{3} and b_{1}: b_{2}: b_{3} are perpendicular, then a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}=0
Explanation:
B(0,-1,3) is a point on the line.
Therefore, \vec{A}=-\hat{j}+\frac{\pi}{3} \hat{k}
Also direction ratios of the line are (0-2):(-1+3):(3+1)

Any point on the line will be of the form (-\lambda, \lambda-1,2 \lambda+3)
So the foot of the perpendicular is of the form (-\lambda, \lambda-1,2 \lambda+3)
The direction ratios of the perpendicular is

From the direction ratio of the line and the direction ratio of its perpendicular, we have
So, the foot of the perpendicular is \left(\frac{-5}{3}, \frac{2}{3}, \frac{19}{3}\right)

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