JEE Mains · Maths · STD 12 - 11. three dimension geometry
The plane passing through the poin \((4, -1, 2)\) and parallel to the lines \(\frac{{x + 2}}{3} = \frac{{y - 2}}{{ - 1}} = \frac{{z + 1}}{2}\) and \(\frac{{x - 2}}{1} = \frac{{y - 3}}{2} = \frac{{z - 4}}{3}\) also passes through the point
- A \((1, 1, -1)\)
- B \((1, 1, 1)\)
- C \((-1,-1,-1)\)
- D \((-1,-1,1)\)
Answer & Solution
Correct Answer
(B) \((1, 1, 1)\)
Step-by-step Solution
Detailed explanation
\(\vec n = {\vec n_1} \times {\vec n_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 3&{ - 1}&2\\ 1&2&3 \end{array}} \right| = 7\hat i - 7\hat j + 7\hat k\) Equation of plane is \(-7(x-4)-7(y+1)+7(z-2)=0\) \(\Rightarrow-7 x-7 y+7 z+7=0 \Rightarrow x+y-z=1\)
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