A) (1, 1, -1)
B) (1, 1, 1)
C) (-1, -1, -I)
D) (-1, -1, 1)
Correct Answer: B
Solution :
Normal of plane \[=\,\,\,\left| \begin{matrix} \widehat{i} \\ 3 \\ 1 \\ \end{matrix}\,\,\,\,\,\,\begin{matrix} \widehat{j} \\ -1 \\ 2 \\ \end{matrix}\,\,\,\,\,\,\begin{matrix} \widehat{k} \\ 2 \\ 3 \\ \end{matrix} \right|\,\,\,\,\,\] \[=\,\,\,\,\widehat{i}(-3-4)-\widehat{j}(9-2)+\widehat{k}(6+1)\] \[=\,\,-7\widehat{i}\,-7\widehat{j}\,+\,7\widehat{k}\] D.r.s of normal of plane = 1, 1, -1 Eq. of plane \[1(x-4)+1(y+1)-1(z-2)=0\] \[x+y-z-1=0\] Option point (1, 1, 1) satisfy itYou need to login to perform this action.
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