JEE Mains · Maths · STD 12 - 11. three dimension geometry
The coordinates of the foot of the perpendicular from the point \((1, -2, 1)\) on the plane containing the lines, \(\frac{{x + 1}}{6} = \frac{{y - 1}}{7} = \frac{{z - 3}}{8}\) and \(\frac{{x - 1}}{3} = \frac{{y - 2}}{5} - \frac{{z - 3}}{7}\) is
- A \((2, -4, 2)\)
- B \((-1 , 2, -1)\)
- C \((0, 0, 0)\)
- D \((1, 1, 1)\)
Answer & Solution
Correct Answer
(C) \((0, 0, 0)\)
Step-by-step Solution
Detailed explanation
\(\vec n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 6&7&8\\ 3&5&7 \end{array}} \right|\) \({=(9,-18,9)} \) \({=(1,-2,1)}\) \(\therefore\) Equation of plane is \(1(x+1)-2(y-1)+(z-3)=0\)…
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