Question

Find the foot of perpendicular from the point 2i -j +5k on the line r=11i -2j-8k +a(10i-4j-11k).Also find the length of perpendicular

Answer 1

Please see the attachment

Answer 2

The coordinate of foot of perpendicular is ($-\frac{857}{237}$,$\frac{11}{237}$,$\frac{-29}{237}$)

The length of the perpendicular is= 7.67units

Step-by-step explanation:

Given equation of straight line is

r=11$\hat{i}$ - 2$\hat{j}$ - 8$\hat{k}$+a(10$\hat{i}$ - 4$\hat{j}$ -11$\hat{k}$)

Any point on the straight line be (10r-11,-4r+2,-11r+8)

Let the coordinate  of foot of perpendicular be P (10r-11,-4r+2,-11r+8)

Given point is A (2,-1,5)

Therefore the direction ratio of AP Straight line is (10r-11-2,-4r+2+1,-11r+8-5)

=(10r-13,-4r+3,-11r+3)

Since the given line perpendicular to the AP straight line

∴10(10r-13)-4(-4r+3)-11(-11r+3)=0

⇒237r = 175

⇒r=$\frac{175}{237}$

The coordinate of foot of perpendicular is ($-\frac{857}{237}$,$\frac{11}{237}$,$\frac{-29}{237}$)

The length of the perpendicular is=$\sqrt{(2+\frac{857}{237}) ^{2}+(-1-\frac{11}{237} )^{2}+(5+\frac{29}{237}) ^{2} }$ =7.67units