Let a line L pass through the point P(2,3,1) and be parallel to the line x+3 y-2 z-2=0=x-y+2 z. If the distance of L from the point (5,3,8) is alpha, then 3 alpha^2 is equal to :

L:x - 2 / 1 = y - 3 / - 1 = z - 1 / - 1 = lambda Any point on L can be taken as B(lambda + 2, - lambda + 3, - lambda + 1) Let A(5,3,8) So, AB . (hat i - hat j - hat k) = 0 [(lambda - 3)hat i - lambda hat j - (lambda + 7)hat k] . [hat i - hat j - hat k] = 0 lambda - 3 + lambda + lambda + 7 = 0 Therefore lambda = - 4 / 3 -> AB = 13 / 3hat i + 4 / 3hat i - 17 / 3hat k |-> AB | = sqrt 169 / 9 + 16 / 9 + 289 / 9 = sqrt 474 / 3 = alpha 3alpha ^2 = 474 / 9 x 3 = 158

Let a line L pass through the point P231 and be paralle... | null (Mathematics)

L:{{x - 2} \over 1} = {{y - 3} \over { - 1}} = {{z - 1} \over { - 1}} = \lambda

Any point on L can be taken as

B(\lambda + 2, - \lambda + 3, - \lambda + 1)

Let

A(5,3,8)

So,

AB\,.\,(\widehat i - \widehat j - \widehat k) = 0

[(\lambda - 3)\widehat i - \lambda \widehat j - (\lambda + 7)\widehat k]\,.\,[\widehat i - \widehat j - \widehat k] = 0

\lambda - 3 + \lambda + \lambda + 7 = 0

\therefore

\lambda = {{ - 4} \over 3}

\overrightarrow {AB} = {{13} \over 3}\widehat i + {4 \over 3}\widehat i - {{17} \over 3}\widehat k

|\overrightarrow {AB} | = \sqrt {{{169} \over 9} + {{16} \over 9} + {{289} \over 9}}

= {{\sqrt {474} } \over 3} = \alpha

3{\alpha ^2} = {{474} \over 9} \times 3 = 158

Explanation

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