JEE Mains · Maths · STD 12 - 11. three dimension geometry
The line of intersection of the planes \(\vec r.\left( {3\hat i - \hat j + \hat k} \right) = 1\) and \(\vec r.\left( {\hat i + 4\hat j - 2\hat k} \right) = 2\), is
- A \(\frac{{x - \frac{4}{7}}}{{ - 2}} = \frac{y}{7} = \frac{{z - \frac{5}{7}}}{{13}}\)
- B \(\frac{{x - \frac{4}{7}}}{2} = \frac{y}{{ - 7}} = \frac{{z + \frac{5}{7}}}{{13}}\)
- C \(\frac{{x - \frac{6}{{13}}}}{2} = \frac{{y - \frac{5}{{13}}}}{{ - 7}} = \frac{z}{{ - 13}}\)
- D \(\frac{{x - \frac{6}{{13}}}}{2} = \frac{{y - \frac{5}{{13}}}}{7} = \frac{z}{{ - 13}}\)
Answer & Solution
Correct Answer
(C) \(\frac{{x - \frac{6}{{13}}}}{2} = \frac{{y - \frac{5}{{13}}}}{{ - 7}} = \frac{z}{{ - 13}}\)
Step-by-step Solution
Detailed explanation
\(\vec n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} \) \( \Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 3&{ - 1}&1\\ 1&4&{ - 2} \end{array}} \right| = \hat i\left( { - 2} \right) - \hat j\left( { - 7} \right) + \hat k\left( {13} \right)\)…
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