Find the distance the point P(0,-2,-7) is to the line through the two points Q(1

Question

Find the distance the point P(0,-2,-7) is to the line through the two points
Q(1,-1,-4), and R(-2,-3,-5).

Explanation / Answer

Line: < -3 , 3 , -5 > + t < -4 - -3 , 0 - 3 , -7 - -5 > < -3 , 3 , -5 > + t < -4 + 3 , 0 - 3 , -7 + 5 > < -3 , 3 , -5 > + t < -1 , - 3 , -2 > Line is: x = -3 - t as d + g t y = 3 - 3t as e + h t z = -5 -2t as f + i t Point (-2 , 5 , -9) as (a , b , c ) Distance Formula : v( (a - d)^2 + (b - e)^2 + (c - f)^2 ) * sin( cos^-1( [ (a - d)g + (b - e)h + (c - f) i ] / [ v( (a - d)^2 + (b - e)^2 + (c - f)^2 ) v( g^2 + h^2 + i^2 ) ] ) ) v( (-2 - -3)^2 + (5- 3)^2 + (-9 - -5)^2 ) * sin( cos^-1( [ (-2 - -3)*-1 + (5- 3)*-3 + (-9 - -5)*-2 ] / [ v( (-2 - -3)^2 + (5- 3)^2 + (-9 - -5)^2 ) v( (-1)^2 + (-3)^2 + (-2)^2 ) ] ) ) v( (-2 + 3)^2 + 2^2 + (-9 + 5)^2 ) * sin( cos^-1( [ (-2 + 3)*-1 + 2 * -3 + (-9 + 5)*-2 ] / [ v( (-2 + 3)^2 + 2^2 + (-9 + 5)^2 ) v( 1 + 9 + 4 ) ] ) ) v( 1^2 + 4 + (-4)^2 ) * sin( cos^-1( [ 1*-1 - 6 + -4*-2 ] / [ v( 1^2 + 4 + (-4)^2 ) v(14) ] ) ) v( 1 + 4 + 16 ) * sin( cos^-1( [ -1 - 6 + 8 ] / [ v( 1 + 4 + 16 ) v(14) ] ) ) v(21) * sin( cos^-1( 1 / [ v(21) v(14) ] ) ) v(21) * sin( cos^-1( 1 / 7v(6) ) ) v(129/7)

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