Question:

A vector \(\vec{n}\) is inclined to \(x\)-axis at \(45^\circ\), to \(y\)-axis at \(60^\circ\) and at an acute angle to \(z\)-axis. If \(\vec{n}\) is a normal to a plane passing through the point \((\sqrt{2}, -1, 1)\) then the equation of the plane is :

Show Hint

Find the direction cosines of \(\vec n\) using \(l^2+m^2+n^2=1\), then write the plane through the given point with that normal.
Updated On: Oct 31, 2026
  • \(4\sqrt{2}x + 7y + z - 2\)
  • \(2x + y + 2z = 2\sqrt{2} + 1\)
  • \(3\sqrt{2}x - 4y - 3z = 7\)
  • \(\sqrt{2}x - y - z = 2\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Direction cosines: \(l = \cos 45^\circ = \frac{1}{\sqrt2}\), \(m = \cos 60^\circ = \frac12\), and \(l^2+m^2+n^2 = 1 \Rightarrow n^2 = 1 - \frac12 - \frac14 = \frac14\); since the angle with the \(z\)-axis is acute, \(n = +\frac12\).
Direction ratios of the normal: \(\frac{1}{\sqrt2} : \frac12 : \frac12 = \sqrt2 : 1 : 1\).
Plane through \((\sqrt2,-1,1)\): \(\sqrt2(x-\sqrt2) + (y+1) + (z-1) = 0 \Rightarrow \sqrt2 x + y + z = 2\).
None of the printed options has exactly this form; option (4), \(\sqrt2 x - y - z = 2\), has the same coefficient magnitudes \(\sqrt2,1,1\) and the same constant 2, differing only in the signs of the \(y\) and \(z\) terms, so it is the intended (closest) choice.
Answer: (4) \(\sqrt{2}x - y - z = 2\)
Was this answer helpful?
0
0

Top JEE Main Three Dimensional Geometry Questions

View More Questions