Question 284 of 521MediumMCQ
If for a>0, the feet of perpendiculars from the points A(a,−2a,3) and B(0,4,5) on the plane lx+my+nz=0 are the points C(0,−a,−1) and D respectively, then the length of line segment CD is equal to:
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Official solution
Let \[\phi \] is the angle between \[\overrightarrow {AB} \] and \[\overrightarrow n \].
CD = AR =, AB |sin\[\phi\]
CD =, AB, \[\sqrt {1 - {{\cos }^2}\phi } \]
CD =, AB, \[\sqrt {1 - {{\left( {{{\overrightarrow {AB} .\,\overrightarrow n } \over {|AB|}}} \right)}^2}} \]
\[ = \sqrt {{{(AB)}^2} - {{(\overrightarrow {AB} \,.\,\overrightarrow n )}^2}} \]
[ \[\cos \phi = {{\overrightarrow {AB} \,.\,\overrightarrow n } \over {|\overrightarrow n ||\overrightarrow {AB} |}}\]]
\[|\overrightarrow {AB} |\, = a\widehat i - (2a + 4)\widehat j - 2\widehat k\]
\[\overrightarrow {AB} \,.\,\overrightarrow n = la - (2a + 4) - 2n\]
C on plane
(0)l - am - n = 0 ..... (1)
Also, \[\overrightarrow {AC} \] || \[\overrightarrow n \]
\[{a \over l} = {{ - a} \over m} = {4 \over n}\]
m = -l \& an + 4m = 0 ..... (2)
From (1) and (2)
a2m + an = 0
\[\underline {4m + an = 0} \]
(a2 - 4)m = 0 \[ \Rightarrow \] a = 2
2m + n = 0 .... (1)
m + l = 0
l2 + m2 + n2 = 1
m2 + m2 + 4m2 = 1
m2 = \[{1 \over 6}\]
m = \[{1 \over {\sqrt 6 }}\]
n = \[{{ - 2} \over {\sqrt 6 }}\]
l = \[{{ - 1} \over {\sqrt 6 }}\]
Now, \[\overrightarrow {AB} \,.\,\overrightarrow n = 2\left( {{{ - 1} \over {\sqrt 6 }}} \right) - 8\left( {{{ - 1} \over {\sqrt 6 }}} \right) - 2\left( {{{ - 2} \over {\sqrt 6 }}} \right)\]
\[ = {{ - 2 - 8 + 4} \over {\sqrt 6 }} = - \sqrt 6 \]
\[|\overrightarrow {AB} |\, = \sqrt {4 + 64 + 4} = \sqrt {72} \]
\[CD = \sqrt {72 - 6} \]
\[CD = \sqrt {66} \]
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