Question 274 of 521MediumMCQ
A plane passes through the points A(1,2,3), B(2,3,1), C(2,4,2). If O is the origin and P is (2,−1,1), then the projection of OP on this plane is of length:
Solutions · 1
Official solution
A(1, 2, 3), B(2, 3, 1), C(2, 4, 2), O(0, 0, 0)
Equation of plane passing through A, B, C will be
\[\left| {x−12−12−1y−23−24−2z−31−32−3 } \right| = 0\]
\[ \Rightarrow \left| {x−111y−212z−3−2−1 } \right| = 0\]
\[ \Rightarrow (x - 1)( - 1 + 4) - (y - 2)( - 1 + 2) + (z - 3)(2 - 1) = 0\]
\[ \Rightarrow (x - 1)(3) - (y - 2)(1) + (z - 3)(1) = 0\]
\[ \Rightarrow 3x - 3 - y + 2 + z - 3 = 0\]
\[ \Rightarrow 3x - y + z - 4 = 0\], is the required plane.
Now, O(0, 0, 0) \& P(2, -1, 1)
Plane is 3x−y+z−4=0
O' \& P' are foot of perpendiculars.
For O'
\[{{x - 0} \over 3} = {{y - 0} \over { - 1}} = {{z - 0} \over 1} = {{ - (0 - 0 + 0 - 4)} \over {9 + 1 + 1}}\]
\[{x \over 3} = {y \over { - 1}} = {z \over 1} = {4 \over {11}}\]
\[ \Rightarrow O'\left( {{{12} \over {11}},{{ - 4} \over {11}},{4 \over {11}}} \right)\]
for P'
\[{{x - 2} \over 3} = {{y + 1} \over { - 1}} = {{z - 1} \over 1} = {{ - (3(2) - ( - 1) + 1 - 4)} \over {9 + 1 + 1}}\]
\[{{x - 2} \over 3} = {{y + 1} \over { - 1}} = {{z - 1} \over 1} = \left( {{{ - 4} \over {11}}} \right)\]
\[P'\left( {{{ - 12} \over {11}} + 2,{4 \over {11}} - 1,{{ - 4} \over {11}} + 1} \right)\]
\[ \Rightarrow P'\left( {{{10} \over {11}},{{ - 7} \over {11}},{7 \over {11}}} \right)\]
\[O'P' = \sqrt {{{\left( {{{10} \over {11}} - {{12} \over {11}}} \right)}^2} + {{\left( {{{ - 7} \over {11}} + {4 \over {11}}} \right)}^2} + {{\left( {{7 \over {11}} - {4 \over {11}}} \right)}^2}} \]
\[ \Rightarrow O'P' = {1 \over {11}}\sqrt {4 + 9 + 9} \]
\[ \Rightarrow O'P' = {{\sqrt {22} } \over {11}}\]
\[ \Rightarrow O'P' = {{\sqrt 2 \times \sqrt {11} } \over {11}}\]
\[ \Rightarrow O'P' = \sqrt {{2 \over {11}}} \]
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