Question

find the distance of the point (-2, 3, 4) from the line
(x+2)/3 = (2y+3)/4 = (3z+4)/5 measured parallel to the plane
4x+12y-3z +1=0. ​

Answer 1

$\bf\large\underline\green{Answer:-}$

➤ 17/2 = 8.5

$\bf\large\underline\pink{Explanation :-}$

➤Let P = ( -2, 3, -4 ).  Let π be the plane through P parallel to the given plane and let Q be the point where the given line meets π.  The required distance is then the length of the segment PQ.

Step 1 :  Find π

As π is parallel to the given plane, its equation differs only in the constant term.  So the equation for π is

4x + 12y - 3z + c = 0

for some constant c.  Since P is in π, its coordinates satisfy the equation, so

4(-2) + 12(3) - 3(-4) + c = 0  =>  -8 + 36 + 12 + c = 0  =>  c = -40

Therefore the equation for π is

4x + 12y - 3z - 40 = 0       ... (1)

Step 2 :  Find Q

For points on the line, (x+2)/3=(2y+3)/4  =>  4x+8 = 6y+9  =>  6y = 4x-1

and (x+2)/3=(3z+4)/5  =>  5x+10 = 9z+12  =>  9z = 5x-2

For points in plane π,

4x + 12y - 3z - 40 = 0  =>  12x + 36y - 9z - 120 = 0

So for point Q which is both on the line and in π,

12x + 36y - 9z - 120 = 0

=> 12x + 6(4x-1) - (5x-2) - 120 = 0

=> 12x + 24x - 6 -5x + 2 - 120 = 0

=> 31x = 124

=> x = 4

Then y = (4x-1)/6 = 15/6 = 5/2  and  z = (5x-2)/9 = 18/9 = 2.

Thus Q = ( 4, 5/2, 2 ).

Step 3 :  Length of PQ

We have P = ( -2, 3, -4 ) and Q = ( 4, 5/2, 2 ).

The required distance is the length of PQ, which is

√( (-2-4)² + (3-5/2)² + (-4-2)² )

= √( (-6)² + (1/2)² + (-6)² )

= √( 36 + 1/4 + 36 )

= √( 72 + 1/4 )

= √( 288/4  +  1/4 )

= √( 289 / 4 )

= 17 / 2

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