JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the image of the point \(\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)\) in the plane \(x-2 y+z-2=0\) be \(P\). If the distance of the point \(Q(6,-2, \alpha), \alpha > 0\), from \(P\) is 13 , then \(\alpha\) is equal to \(...........\).
- A \(14\)
- B \(13\)
- C \(15\)
- D \(12\)
Answer & Solution
Correct Answer
(C) \(15\)
Step-by-step Solution
Detailed explanation
Let the image point be \(P(x_p, y_p, z_p)\). The formula for the image of a point \((x_1, y_1, z_1)\) in the plane \(ax+by+cz+d=0\) is \(\frac{x_p-x_1}{a} = \frac{y_p-y_1}{b} = \frac{z_p-z_1}{c} = k\), where \(k = -2\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}\).…
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