Question

A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\prime}$, to the $y$-axis at $45$ 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 $\vec{v}$, then

• A. $\sqrt{2} a-b+c=1$
• B. $a+\sqrt{2} b+c=1$
• C. $\sqrt{2} a+b+c=1$
• D. $a+b+\sqrt{2} c=1$

Answer 1

The Correct Option is B

To solve this problem, let's start by understanding the given conditions and how to use them to find the equation of the plane. We are given a vector \(\vec{v}\)that is inclined to the coordinate axes. Specifically, it is inclined to the x-axis at \(60^\circ\), to the y-axis at \(45^\circ\), and to the z-axis at some acute angle. Since the vector is in the first octant, all direction cosines are positive.

1. Direction cosines, \(l, m, n\), are the cosines of the angles a vector makes with the x, y, and z axes, respectively.
2. From the given inclinations:
   • \(l = \cos 60^\circ = \frac{1}{2}\)
   • \(m = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
3. For direction cosines, it holds that: \(l^2 + m^2 + n^2 = 1\)
4. Substituting known values: \(\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + n^2 = 1\)
5. Simplifying: \(\frac{1}{4} + \frac{1}{2} + n^2 = 1\)
6. Further simplification gives: \(\frac{3}{4} + n^2 = 1 \Rightarrow n^2 = \frac{1}{4} \Rightarrow n = \frac{1}{2}\)(since \(n\)is positive)
7. Thus, the vector \(\vec{v} = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right)\).
8. The plane we are looking for is normal to \(\vec{v}\)and passes through the points \((\sqrt{2}, -1, 1)\)and \((a, b, c)\).
9. If a plane is normal to vector \(\vec{v}\), then: \(\frac{1}{2}x + \frac{1}{\sqrt{2}}y + \frac{1}{2}z = d\), where \(d\)is a constant.
10. Substitute the point \((\sqrt{2}, -1, 1)\)into the plane equation: \(\frac{1}{2}\sqrt{2} + \frac{1}{\sqrt{2}}(-1) + \frac{1}{2}(1) = d\)
11. After substituting and solving: \(\frac{\sqrt{2}}{2} - \frac{1}{\sqrt{2}} + \frac{1}{2} = d\)
12. This simplifies to: \(1 = d\)
13. Hence, the equation of the plane is: \(\frac{1}{2}x + \frac{1}{\sqrt{2}}y + \frac{1}{2}z = 1\)
14. Rewriting the plane equation by substituting in the point \((a, b, c)\): \(\frac{1}{2}a + \frac{1}{\sqrt{2}}b + \frac{1}{2}c = 1\)
15. Multiplying throughout by 2 to clear fractions: \(a + \sqrt{2}b + c = 1\)

Therefore, the correct answer is \(a + \sqrt{2}b + c = 1\).