A) \[3x+2y-3z\,\,=\,\,0\]
B) \[x+2y-2z\,\,=\,\,0\]
C) \[x-2y+z=0\]
D) \[~5x+2y-4z=0\]
Correct Answer: C
Solution :
Vector \[\bot \] to given plane = \[\left| \begin{align} & \widehat{i}\,\,\,\,\,\,\,\widehat{j}\,\,\,\,\,\,\widehat{k} \\ & 3\,\,\,\,4\,\,\,\,\,\,2 \\ & 4\,\,\,\,\,2\,\,\,\,\,3 \\ \end{align} \right|\] \[=\text{ }\widehat{i}\left( 12-4 \right)-\widehat{j}\left( 9-8 \right)+\widehat{k}\left( 6-16 \right)\] \[=\,\,\,8\widehat{i}-\widehat{j}-10\widehat{k}~~~~~~~~~...\left( 1 \right)\] Vector parallel to given line \[=\text{ }2\widehat{i}+3\widehat{j}+4\widehat{k}\] ... (2) Vector \[\bot \] to both (1) & (2) vector \[=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 8 & -1 & -10 \\ 2 & 3 & 4 \\ \end{matrix} \right|\] \[=\,\,\,\widehat{i}\left( -4+30 \right)-\widehat{j}\left( 32+20 \right)+\widehat{k}\left( 24+2 \right)\] \[=26\widehat{i}-52\widehat{j}+26\widehat{k}\] Dr?s of normal of required plane is \[(26,-\,52,\]\[26)\Rightarrow \left( 1,-2,\,\,1 \right)\] Equation of plane whose Dr?s of Normal is \[\left( 1,-2,\text{ }1 \right)\] and passes through origin \[1.\,\left( x-0 \right)-2\left( y-0 \right)+1.\left( z-0 \right)=0\] \[x-2y+z=0\]
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