The length of the projection of the line segment joining the points (5,-1,4) and (4,-1,3) …

Question

The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is:

Accepted Answer

PQ  is the projection of line segment AB  on the plane x + y + z = 7

P and Q are called foot of perpendicular on the plane x + y + z = 7

Let P = (x, y, z) then

$${{x - 5} \over 1} = {{y + 1} \over 1} = {{z - 4} \over 1} = {{ - \left( {5 - 1 + 4} \right)} \over {{1^2} + {1^2} + {1^2}}}$$

$$ \Rightarrow \,\,\,\,x - 5 = y + 1 = z - 4 =  - {8 \over 3}$$

$$\therefore\,\,\,$$  x = $${7 \over 3}$$ ,  y = - $${{11} \over 3},$$ z = $${4 \over 3}$$

$$\therefore\,\,\,$$ Point P = $$\left( {{7 \over 3}, - {{11} \over 3},{4 \over 3}} \right)$$

Let Q = (x1, y1, z1) then

$${{{a_1} - 4} \over 1} = {{{y_1} + 1} \over 1} = {{{z_1} - 3} \over 1} = {{ - \left( {4 - 1 + 3} \right)} \over {{1^2} + {1^2} + {1^2}}}$$

$$ \Rightarrow \,\,\,\,$$ x1 - 4 = y1 +1= z1 - 3 = - 2

$$\therefore\,\,\,$$  x = 2, y = - 3,  z = 1

$$\therefore\,\,\,$$  Point Q = (2, - 3, 1)

Now length of PQ is

$$\sqrt {{{\left( {{7 \over 3} - 2} \right)}^2} + {{\left( { - {{11} \over 3} + 3} \right)}^2} + {{\left( {{4 \over 3} - 1} \right)}^2}} $$

$$ = \sqrt {{1 \over 9} + {4 \over 9} + {1 \over 9}} $$

$$ = \sqrt {{6 \over 9}} $$

$$ = \sqrt {{2 \over 3}} $$

Solutions · 1