Question

foot of perpendicular from origin to line x+2y+3z+4=0=2x+3y+4z+5

Answer 1

Given:

$\to x + 2y + 3z + 4 = 0 \\\\\to 2x + 3y + 4z + 5 = 0$

To Find:

The foot of the perpendicular from origin to the line=?

Solution:

In order to achieve the equation of the line between these two planes,

$\to x + 2y + 3z + 4 = 0 \\\\\to 2x + 3y + 4z + 5 = 0$

If the variable Z is free then the line fur of the perpendicular variable Z = t is free.

$\to x + 2y = -3t - 4 - (1) \\\\\to 2x + 3y = -4t - 5 - (2)$

Solve the value x and y in terms of t:

$\to 2x + 4y = -6t - 8\\\\\to 2x + 3y = -4t - 5\\\\\to y = -2t -3\\\\\to x = -3t - 4 - 2 ( -2t - 3 )\\\\\to x = -3t - 4 +4t + 6\\\\\to x = t + 2$

parametrizing the line:

$\to (t + 2 )i + (-2t - 3 )j + tk$

Line equation $= 2i -3j + t ( i -2j +k)$

When the parameterizing line is the foot of the perpendicular,  Calculating the Dot product and parallel to  = 0 .

$\to (( t + 2 )i + (-2t - 3 )j + tk ).( i - 2j + k) = 0\\\\\to t + 2 + 4t + 6 + t = 0\\\\\to 6t = -8\\\\\to t = \frac{-8}{6} \\\\\to t= \frac{-4}{3}$

So, the points are​ $(\frac{2}{3} , \frac{-1}{3}, \ and\ \frac{-4}{3})$

Answer 2

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