JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the point \(\left( {2,\alpha ,\beta } \right)\) lies on the plane which passes through the points \((3, 4, 2)\) and \((7, 0, 6)\) and is perpendicular to the plane \(2x - 5y = 15\) , then is equal to \({2\alpha - 3\beta }\) is equal to
- A \(12\)
- B \(7\)
- C \(5\)
- D \(17\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
\(A(7,0,6)\) and \(B(3,4,2)\) \(\overrightarrow {{\rm{AB}}} = - 4\widehat {\rm{i}} + 4\widehat {\rm{j}} - 4\widehat {\rm{k}}\) Also \(2 \hat{i}-5 \hat{j}\) is parallel to the plane \(\Rightarrow\) Normal perpendicular to the required plane is \(\)\left| {\begin{array}{*{20}{c}}…
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