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Find the equation of the plane passing through the line of intersection of the planes \(\vec r .(\hat i + \hat j + \hat k) = 10 \) and \(\vec r .(2\hat i +3 \hat j - \hat k) +4 = 0 \) and passing through the point (-2, 3, 1). |
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Answer» Equation of plane passing through the line of intersection of given plane is \(\vec r .(\hat i + \hat j + \hat k) + \lambda (2\hat i + 3\hat j - \hat k)= 10 -4 \lambda\) where \(\vec r = x \hat i + y\hat j +z \hat k\) ⇒ \((x \hat i + y\hat j + z\hat k).((1 + 2\lambda)\hat i + (1 + 3\lambda)\hat j + (1 - \lambda)\hat k) = 10 - 4\lambda\) .....(i) It is passing through point (-2, 3, 1). \(\therefore -2 ( 1 + 2\lambda )+ 3(1+ 3\lambda)+ 1(1 - \lambda) = 10 - 4\lambda\) ⇒ \(- 2-4 \lambda + 3 + 9\lambda + 1 - \lambda = 10 - 4 \lambda\) ⇒ \(8 \lambda = 10 -2\) ⇒ \(8 \lambda = 8\) ⇒ \(\lambda = 1\) Hence, the equation of required plane is \(\vec r .(3 \hat i + 4 \hat j) = 6\) (From(i)) or \(3x + 4y = 6\) |
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