A plane passing through the point (3,1,1) contains two lines whose direction ratios are 1,−2,2 and 2,3,−1 respectively. If this plane also passes through the point (α,−3,5), then α is equal to:
Solutions · 1
Official solution
As normal is perpendicular to both the lines so normal vector to the plane is
\[\overrightarrow n = \left( {\widehat i - 2\widehat j + 2\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)\]
\[\overrightarrow n = \left| {i12j−23k2−1 } \right|\]
\[\overrightarrow n = \left( {2 - 6} \right)\widehat i - \left( { - 1 - 4} \right)\widehat j + \left( {3 + 4} \right)\widehat k\]
\[\overrightarrow n = - 4\widehat i + 5\widehat j + 7\widehat k\]
Now equation of plane passing through (3,1,1) is
\[ \Rightarrow \] --4(x -- 3) + 5(y -- 1) + 7(z -- 1) = 0
\[ \Rightarrow \] --4x + 12 + 5y -- 5 + 7z -- 7 = 0
\[ \Rightarrow \] --4x + 5y + 7z = 0 ...(1)
Plane is also passing through (\[\alpha \], --3, 5) so this point satisfies the equation of plane so put in equation (1)
--4\[\alpha \] + 5 \times (--3) + 7 \times (5) = 0
\[ \Rightarrow \] --4\[\alpha \] -- 15 + 35 = 0
\[ \Rightarrow\alpha \] = 5