A vector v in the first octant is inclined to the x-axis at 60°, to the y-axis at 45° and to the z-a

Q: Why do we reject the negative value for $\cos \gamma$?

The problem explicitly states that $\gamma$ is an acute angle. In the range $[0, \pi/2]$, the cosine function is strictly non-negative. If we chose the negative root, we would be describing a vector pointing into a different octant, violating the problem's premise. Always check your constraints!

Q: Can I use any vector proportional to the direction cosines as the normal vector?

Absolutely! The equation of a plane $Ax + By + Cz = D$ is defined by the ratio of the coefficients. Scaling the normal vector by a constant factor does not change the orientation of the plane, only the magnitude of the normal vector itself. Using $(1, \sqrt{2}, 1)$ is mathematically equivalent to using $(1/2, 1/\sqrt{2}, 1/2)$.

Q: Why is the plane equation $x + \sqrt{2}y + z = 1$?

This comes from the point-normal form. When you substitute the point $(\sqrt{2}, -1, 1)$ into the equation $1(x - \sqrt{2}) + \sqrt{2}(y - (-1)) + 1(z - 1) = 0$, the constant terms simplify beautifully. The $-\sqrt{2}$ and $+\sqrt{2}$ cancel out, leaving you with the elegant form $x + \sqrt{2}y + z = 1$.

Question

A vector $\vec{v}$ 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 $\vec{v}$, then

Options:

(A) $\sqrt{2}a + b + c = 1$

(B) $a + b + \sqrt{2}c = 1$

(C) $a + \sqrt{2}b + c = 1$

(D) $\sqrt{2}a - b + c = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are not just solving a problem; we are embarking on a journey through 3D space. Imagine yourself standing at the origin of a Cartesian coordinate system, looking at a vector $\vec{v}$ floating in the first octant.

This vector is defined by its orientation relative to the three fundamental axes: $x$, $y$, and $z$. The problem provides two angles: $60^\circ$ with the $x$-axis and $45^\circ$ with the $y$-axis.

### The Master Key: Direction Cosines

In the world of vectors, we have a fundamental identity that acts as our North Star. For any vector, the sum of the squares of the cosines of the angles it makes with the coordinate axes is always unity.

We express this as:

$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$

This is the Pythagorean theorem extended into three dimensions, ensuring the vector's projection onto the unit sphere always lands on the surface. Substituting our known values, $\alpha = 60^\circ$ and $\beta = 45^\circ$, we obtain:

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

Simplifying this, we find:

$$\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1 \Rightarrow \frac{3}{4} + \cos^2 \gamma = 1$$

Thus, $\cos^2 \gamma = \frac{1}{4}$. Taking the square root gives us $\pm \frac{1}{2}$.

Since the problem specifies that $\gamma$ is an acute angle, we know the cosine must be positive. Therefore, we discard the negative root and conclude $\cos \gamma = \frac{1}{2}$.

### Constructing the Plane: The Soul of the Normal Vector

With our direction cosines identified as $(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})$, we have defined the orientation of our plane. The direction cosines of a vector serve as the direction ratios of its normal vector.

To simplify our calculations, we scale these ratios by $2$ to obtain the normal vector $\vec{n} = (1, \sqrt{2}, 1)$. We now invoke the point-normal form of a plane equation:

$$A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$$

Given the point $P(\sqrt{2}, -1, 1)$ lies on this plane, we substitute our normal vector and point coordinates:

$$1(x - \sqrt{2}) + \sqrt{2}(y - (-1)) + 1(z - 1) = 0$$

### The Elegance of Cancellation

Expanding this equation, we observe the following:

$$x - \sqrt{2} + \sqrt{2}y + \sqrt{2} + z - 1 = 0$$

The terms $-\sqrt{2}$ and $+\sqrt{2}$ cancel out perfectly. This leaves us with the pristine equation of the plane:

$$x + \sqrt{2}y + z = 1$$

Finally, we are told that another point $Q(a, b, c)$ lies on this plane. By substituting these coordinates into our derived equation, we arrive at the final relation:

**$a + \sqrt{2}b + c = 1$**

Select Answer:

Visualized Solution

Visualizing the Vector $v$

The Direction Cosine Identity

Substituting Known Angles

Evaluating the Squares

Solving for $cos^{2} γ$

Determining $cos γ$

Defining the Normal Vector $v$

Equation of a Plane

Substituting the Values

Expanding the Equation

Final Plane Equation

Substituting Point $Q (a, b, c)$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Key

Constructing the Plane

The Elegance of Cancellation

Similar Questions

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

The equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis is:

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

The vector equation of the plane passing through the intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (\hat{i} - 2 \hat{j}) = - 2$ , and the point $(1, 0, 2)$ is :

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is:

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then $3 (α + β + γ)$ equals

Select Answer:

Visualized Solution

Visualizing the Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

The Direction Cosine Identity

Substituting Known Angles

Evaluating the Squares

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2γ

Determining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosγ

Defining the Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v