If the line of intersection of the planes ax + by = 3 and ax + by + cz = 0, a 0 makes an angle 30°

Q: Wait, why do we use $\sin \theta$ here? I thought the angle between vectors always involved $\cos \theta$?

That is a brilliant catch! You are absolutely right that the dot product formula $\vec{A} \cdot \vec{B} = |A||B| \cos \theta$ gives the cosine of the angle between two vectors. However, here we are finding the angle between a *line* and a *plane*. The angle between a line and a plane is the complement of the angle between the line and the *normal* to the plane. Since $\cos(90^\circ - \theta) = \sin \theta$, we switch to sine. Always visualize the normal vector when dealing with planes!

Q: Can I just ignore the $c$ in the direction ratios? Is that mathematically legal?

Yes, it is! Remember that direction ratios are just proportional components. If a vector is $(bc, -ac, 0)$, it points in the exact same direction as $(b, -a, 0)$ because we are just scaling the vector by a factor of $1/c$. As long as $c \neq 0$, this simplification is perfectly valid and makes your algebra much cleaner.

Q: The problem gives $a > 0$. Does that change the final direction cosines?

It acts as a constraint to help us define the direction uniquely. When we solve $b^2 = a^2$, we get $b = \pm a$. The condition $a > 0$ ensures we are looking at a specific orientation. Since the direction cosines are ratios, the sign of $a$ will cancel out in the normalization process, but it helps us maintain consistency in our vector components.

Question

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0, a > 0$ makes an angle $30^\circ$ with the plane $y - z + 2 = 0$, then the direction cosines of the line are :

Options:

(A) $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0$

(B) $\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0$

(C) $\frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}}, 0$

(D) $\frac{1}{2}, -\frac{\sqrt{3}}{2}, 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The intersection of two planes, $P_1$ and $P_2$, forms a line $L$. To find the direction of this line, we must identify a vector $\vec{v}$ that is perpendicular to the normal vectors of both planes, $\vec{n}_1$ and $\vec{n}_2$.

The normal vectors are defined as:

$$\vec{n}_1 = a\hat{i} + b\hat{j} + 0\hat{k}$$
$$\vec{n}_2 = a\hat{i} + b\hat{j} + c\hat{k}$$

### The Birth of the Direction Vector

We utilize the cross product to determine the direction vector $\vec{v}$ of the line $L$:

$$\vec{v} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a & b & 0 \ a & b & c \end{vmatrix}$$

Expanding the determinant, we obtain:

$$\vec{v} = \hat{i}(bc - 0) - \hat{j}(ac - 0) + \hat{k}(ab - ab)$$
$$\vec{v} = bc\hat{i} - ac\hat{j} + 0\hat{k}$$

Since direction ratios are proportional, we simplify the vector by dividing by $c$ (assuming $c 
eq 0$):

$$\vec{v} = (b, -a, 0)$$

### The Dance with the Third Plane

We introduce the third plane, $P_3: y - z + 2 = 0$, which has a normal vector $\vec{n}_3 = (0, 1, -1)$. The line $L$ makes an angle of $30^{\circ}$ with this plane.

The angle $	heta$ between a line with direction vector $\vec{v}$ and a plane with normal $\vec{n}$ is given by the sine relationship:

$$\sin 	heta = \frac{| \vec{v} \cdot \vec{n}_3 |}{| \vec{v} | | \vec{n}_3 |}$$

Substituting $	heta = 30^{\circ}$, $\vec{v} = (b, -a, 0)$, and $\vec{n}_3 = (0, 1, -1)$:

$$\sin 30^{\circ} = \frac{1}{2} = \frac{| (b)(0) + (-a)(1) + (0)(-1) |}{\sqrt{a^2 + b^2} \sqrt{0^2 + 1^2 + (-1)^2}}$$

This simplifies to:

$$\frac{1}{2} = \frac{| -a |}{\sqrt{a^2 + b^2} \sqrt{2}}$$

### Final Calculation

To solve for the relationship between $a$ and $b$, we square both sides:

$$\frac{1}{4} = \frac{a^2}{2(a^2 + b^2)}$$

Cross-multiplying yields $2(a^2 + b^2) = 4a^2$, which simplifies to $2b^2 = 2a^2$, or $b^2 = a^2$. This implies **$b = \pm a$**.

Taking $b = a$, the direction vector becomes $(a, -a, 0)$. Normalizing this vector by its magnitude $\sqrt{a^2 + (-a)^2} = a\sqrt{2}$, we find the direction cosines:

$$\left( \frac{a}{a\sqrt{2}}, \frac{-a}{a\sqrt{2}}, 0 ight) = \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 ight)$$

The final direction cosines of the line are **$\left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 ight)$**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the line of intersection of the planes $a x + b y = 3$ and $a x + b y + cz = 0, a > 0$ makes an angle $3 0^{\circ}$ with the plane $y - z + 2 = 0$ , then the direction cosines of the line are :

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

Finding the Direction Vector $v$

Setting up the Cross Product

Calculating the Cross Product

Simplifying Direction Ratios

Angle with the Third Plane

Formula for Angle

Substituting the Values

Evaluating Dot Product and Magnitudes

Squaring Both Sides

Solving for $a$ and $b$

Calculating Direction Cosines

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Birth of the Direction Vector

The Dance with the Third Plane

Final Calculation

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