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JEE Mains · Maths · STD 12 - 11. three dimension geometry
A vector \(\vec n\) is inclined to \(x-\) axis at \(45^o\), to \(y-\) axis at \(60^o\) and at an acute angle to \(z-\) axis. If \(\vec n\) is a normal to a plane passing through the point \(\left( {\sqrt 2 , - 1,1} \right)\) then the equation of the plane is
- A \(4\sqrt 2 x + 7y + z - 2\)
- B \(2x + y + 2z = 2\sqrt 2 + 1\)
- C \(3\sqrt 2 x - 4y - 3z = 7\)
- D \(\sqrt 2 x - y - z = 2\)
Answer & Solution
Correct Answer
(B) \(2x + y + 2z = 2\sqrt 2 + 1\)
Step-by-step Solution
Detailed explanation
Direction cosines of \(\vec{n}\) are \(\frac{1}{2}, \frac{1}{4}, \frac{1}{2}\) Equation of the plane. \(\frac{1}{2}(x-\sqrt{2})+\frac{1}{4}(y+1)+\frac{1}{2}(z-1)=0 \) \(\Rightarrow 2(x-\sqrt{2})+(y+1)+2(z-1)=0 \) \( \Rightarrow 2 x+y+2 z=2 \sqrt{2}-1+2 \)…
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