AP EAMCET · Maths · Three Dimensional Geometry
A plane is making intercepts \(2,3,4\) on \(X, Y\) and \(Z\)-axes respectively. Another plane is passing through the point \((-1,6,2)\) and is perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,4)\). Then angle between the two planes is
- A \(90^{\circ}\)
- B \(\cos ^{-1} \sqrt{\frac{12}{61}}\)
- C \(\cos ^{-1} \sqrt{\frac{11}{61}}\)
- D \(\cos ^{-1} \sqrt{\frac{5}{6}}\)
Answer & Solution
Correct Answer
(C) \(\cos ^{-1} \sqrt{\frac{11}{61}}\)
Step-by-step Solution
Detailed explanation
Given, \(X\)-intercept \((a)=2\) \[ \begin{aligned} & Y \text { - intercept }(b)=3 \\ & Z \text { - intercept }(c)=4 \end{aligned} \] \(\therefore\) Equation of the plane is \(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1\) \[ \begin{aligned} & B=(1,2,3) \\ & C=(-2,3,4) \end{aligned} \]…
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