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 the problem, we start by considering a vector \(\vec{v}\) represented in a three-dimensional space. The vector \(\vec{v}\) is inclined at certain angles to the coordinate axes and lies in the first octant. Let's denote the direction cosines of this vector as \(\text{cos}(\alpha)\), \(\text{cos}(\beta)\), and \(\text{cos}(\gamma)\) corresponding to the x, y, and z axes, respectively.

Given that:

• The angle between the vector \(\vec{v}\) and the x-axis is \(60^\circ\).
• The angle with the y-axis is \(45^\circ\).
• The angle with the z-axis is acute.

We can calculate the direction cosines using the given angles:

• \(\text{cos}(\alpha) = \text{cos}(60^\circ) = \frac{1}{2}\)
• \(\text{cos}(\beta) = \text{cos}(45^\circ) = \frac{1}{\sqrt{2}}\)

Since the sum of the squares of the direction cosines is 1, we have:

\(\text{cos}^2(\alpha) + \text{cos}^2(\beta) + \text{cos}^2(\gamma) = 1\)

Substitute the known values:

\(\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \text{cos}^2(\gamma) = 1\)

Solve for \(\text{cos}^2(\gamma)\):

\(\frac{1}{4} + \frac{1}{2} + \text{cos}^2(\gamma) = 1\)

\(\text{cos}^2(\gamma) = 1 - \frac{3}{4} = \frac{1}{4}\)

Thus, \(\text{cos}(\gamma) = \frac{1}{2}\) or \(-\frac{1}{2}\). Since the angle with the z-axis is acute, \(\text{cos}(\gamma) = \frac{1}{2}\).

The direction ratios of the line representing the vector \(\vec{v}\) are proportional to the direction cosines: \((1/2, 1/\sqrt{2}, 1/2)\). Hence, the direction ratios can be taken as \((1, \sqrt{2}, 1)\).

Now, for the plane passing through the points \((\sqrt{2},-1,1)\) and \((a, b, c)\) which is normal to \(\vec{v}\), we use the plane equation formed using the direction ratios as the normal vector:

The equation of the plane is \(1(x - \sqrt{2}) + \sqrt{2}(y + 1) + 1(z - 1) = 0\), which simplifies to:

\(x + \sqrt{2}y + z = \sqrt{2} - \sqrt{2} + 1 = 1\).

Thus, the correct answer, where the equation holds, is: \(a+\sqrt{2} b+c=1\).