JEE Mains · Maths · STD 12 - 11. three dimension geometry
A vector \(\overrightarrow{ V }\) in the first octant is inclined to the \(x\) axis at \(60^{\circ}\), to the \(y\)-axis at \(45^{\circ}\) and to the z-axis at an acute angle. If a plane passing through the points \((\sqrt{2},-1,1)\) and \(( a , b , c )\), is normal to \(\overrightarrow{ v }\), then
- A \(\sqrt{2} a + b + c =1\)
- B \(a + b +\sqrt{2} c =1\)
- C \(a +\sqrt{2} b + c =1\)
- D \(\sqrt{2} a-b+c=1\)
Answer & Solution
Correct Answer
(C) \(a +\sqrt{2} b + c =1\)
Step-by-step Solution
Detailed explanation
\(\hat{ v }=\cos 60^{\circ} \hat{ i }+\cos 45^{\circ} \hat{ j }+\cos \gamma \hat{ k }\) \(\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^2 \gamma=1 \quad(\gamma \rightarrow \text { Acute })\) \(\Rightarrow \cos \gamma=\frac{1}{2}\) \(\Rightarrow \gamma=60^{\circ}\) Equation of plane…
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