An angle between the plane, x+y+z = 5 and the line of intersection of the planes, 3x+4y+z–1=0 and 5x

Q: Wait, why do we use $\sin \theta$ instead of $\cos \theta$ in the formula?

Great question! The standard dot product formula $\vec{n} \cdot \vec{b} = |\vec{n}| |\vec{b}| \cos \phi$ gives the angle $\phi$ between the line and the normal vector. Since the normal is perpendicular to the plane, the angle between the line and the plane is $\theta = 90^\circ - \phi$. Therefore, $\cos \phi = \cos(90^\circ - \theta) = \sin \theta$. That is why we use sine!

Q: Does the order of the cross product $\vec{n}_2 \times \vec{n}_3$ versus $\vec{n}_3 \times \vec{n}_2$ matter?

Not at all! If you swap the order, you simply get the negative of the direction vector, $-\vec{b}$. Since the angle formula uses the absolute value of the dot product and the magnitude of the vector, the sign cancels out completely. You will get the same result either way.

Q: What if the dot product in the numerator turns out to be zero?

If the dot product is zero, then $\sin \theta = 0$, which means $\theta = 0^\circ$. This tells us that the line is parallel to the plane. It is a perfectly valid geometric scenario, so don't be alarmed if you see it!

Question

An angle between the plane, $x + y + z = 5$ and the line of intersection of the planes, $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$, is

Options:

(A) $\sin^{-1}(3/\sqrt{17})$

(B) $\cos^{-1}(\sqrt{3/17})$

(C) $\sin^{-1}(\sqrt{3/17})$

(D) $\cos^{-1}(3/\sqrt{17})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a vast, three-dimensional room. You have a blue plane, $P_1$, stretching out before you like a floor.

Then, two other planes, $P_2$ and $P_3$, slice through the space like two massive, intersecting walls. Where these two walls meet, they form a sharp, distinct edge—a line of intersection, $L$.

Our mission is to find the angle $	heta$ between this red line $L$ and the blue plane $P_1$. This is not just a calculation; it is a dance of vectors.

### The Secret of the Intersection

To find the angle, we first need to know where the line $L$ is pointing. How do we define a line that lives on two planes simultaneously?

Think about the normal vectors. The normal vector $\vec{n}_2$ is perpendicular to $P_2$, and $\vec{n}_3$ is perpendicular to $P_3$.

Because the line $L$ lies on both planes, it must be perpendicular to both $\vec{n}_2$ and $\vec{n}_3$. This is the perfect setup for the cross product! We define the direction vector $\vec{b}$ of our line as $\vec{b} = \vec{n}_2 	imes \vec{n}_3$.

### Extracting the DNA of the Planes

Let us look at the equations: $P_2: 3x + 4y + z - 1 = 0$ and $P_3: 5x + 8y + 2z + 14 = 0$. The coefficients give us our normal vectors: $\vec{n}_2 = (3, 4, 1)$ and $\vec{n}_3 = (5, 8, 2)$.

Now, we perform the cross product:

$$
\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 1 \ 5 & 8 & 2 \end{vmatrix}
$$

Expanding this, we get $\vec{b} = \hat{i}(4 \cdot 2 - 8 \cdot 1) - \hat{j}(3 \cdot 2 - 5 \cdot 1) + \hat{k}(3 \cdot 8 - 5 \cdot 4)$.

Calculating these values, we find $\vec{b} = \hat{i}(8 - 8) - \hat{j}(6 - 5) + \hat{k}(24 - 20)$, which simplifies to $\vec{b} = (0, -1, 4)$. We have found the direction of our line!

### The Angle of Engagement

Now, we turn our attention to the blue plane $P_1: x + y + z = 5$. Its normal vector is $\vec{n}_1 = (1, 1, 1)$.

We want the angle $	heta$ between the line $L$ (with direction $\vec{b}$) and the plane $P_1$. As we discussed, the angle between the line and the plane is the complement of the angle between the line and the plane's normal.

Thus, we use the formula:

$$
\sin 	heta = \frac{|\vec{n}_1 \cdot \vec{b}|}{|\vec{n}_1| |\vec{b}|}
$$

### The Final Calculation

Let us plug in our values. The dot product $\vec{n}_1 \cdot \vec{b} = (1)(0) + (1)(-1) + (1)(4) = 0 - 1 + 4 = 3$.

The magnitude of the normal is $|\vec{n}_1| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$. The magnitude of the direction vector is $|\vec{b}| = \sqrt{0^2 + (-1)^2 + 4^2} = \sqrt{17}$.

Putting it all together:

$$
\sin 	heta = \frac{|3|}{\sqrt{3} \cdot \sqrt{17}} = \frac{3}{\sqrt{51}}
$$

Since $3 = \sqrt{3} \cdot \sqrt{3}$, we can simplify this to $\sin 	heta = \frac{\sqrt{3}}{\sqrt{17}} = \sqrt{\frac{3}{17}}$.

Therefore, the final angle is:

**$$	heta = \sin^{-1}\left(\sqrt{\frac{3}{17}}ight)$$**

Animated Solution for Mathematics - Three Dimensional Geometry: An angle between the plane, $x + y + z = 5$ and the line of intersection of the planes, $3 x + 4 y + z - 1 = 0$ and $5 x + 8 y + 2 z + 14 = 0$ , is

Select Answer:

Visualized Solution

Visualizing the Geometry

Direction of Line $L$

Setting up the Cross Product

Expanding the Determinant

Direction Vector $b$

Normal Vector of Plane $P_{1}$

Angle Formula: Line and Plane

Substituting the Vectors

Computing the Numerator

Computing the Denominator

Simplifying the Expression

Final Result

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Secret of the Intersection

Extracting the DNA of the Planes

The Angle of Engagement

The Final Calculation

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