A nonzero vector a is parallel to the line of intersection of the plane determined by the vectors i,

Q: Wait, why do we use the cross product of the normals to find the line of intersection?

Great question! Think of it this way: a plane is defined by its normal vector, which is perpendicular to every line lying on that plane. If a line lies on *both* planes, it must be perpendicular to the normal of the first plane AND the normal of the second plane. The only vector that is simultaneously perpendicular to two other vectors is their cross product. It is the most elegant way to find a common direction!

Q: Why do we get two possible answers for the angle?

This is a subtle but crucial point. A vector has a specific direction, but a line of intersection is just a geometric path. It extends in two opposite directions. If vector $\vec{a}$ is the direction of the line, then $-\vec{a}$ is also a valid direction for that same line. When we calculate the angle between a line and another vector, we must consider both directions, which results in two supplementary angles, $\theta$ and $\pi - \theta$.

Q: Can I solve this using the equation of the planes $Ax + By + Cz + D = 0$?

Absolutely! You can find the equation of each plane using the normal vector and a point on the plane. Then, you would have a system of two linear equations with three variables. Solving this system will give you the parametric equation of the line of intersection. However, the cross-product method is generally faster and less prone to algebraic errors in competitive exams like JEE Advanced. Stick to the vector approach whenever possible!

Question

A nonzero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i} + \hat{j}$ and the plane determined by the vectors $\hat{i} - \hat{j}, \hat{i} + \hat{k}$. The angle between $\vec{a}$ and the vector $\hat{i} - 2\hat{j} + 2\hat{k}$ is .........

Sigma Percentile · Accepted Answer

### The Geometry of Intersecting Worlds

Imagine you are standing in a vast, three-dimensional void. You see two flat, infinite sheets—planes—slicing through space. They aren't parallel; they meet, and where they meet, they form a sharp, distinct line of intersection.

This is the heart of our problem. We are given two planes, each defined by a pair of vectors, and we need to find the angle between a vector $\vec{a}$ that runs along this intersection and another vector $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$.

### Phase 1: Unveiling the Normals

To understand the line of intersection, we must first understand the planes themselves. A plane is defined by its normal vector—a vector that stands perfectly upright, perpendicular to the surface.

For the first plane, defined by $\hat{i}$ and $\hat{i} + \hat{j}$, the normal vector $\vec{n}_1$ is the cross product of these two vectors:

$$\vec{n}_1 = \hat{i} 	imes (\hat{i} + \hat{j}) = (\hat{i} 	imes \hat{i}) + (\hat{i} 	imes \hat{j}) = 0 + \hat{k} = \hat{k}$$

It is as simple as that! The first plane is horizontal, lying in the $xy$-plane, with its normal pointing straight up along the $z$-axis.

Now, let us look at the second plane, defined by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$. We calculate its normal $\vec{n}_2$ similarly:

$$\vec{n}_2 = (\hat{i} - \hat{j}) 	imes (\hat{i} + \hat{k}) = (\hat{i} 	imes \hat{i}) + (\hat{i} 	imes \hat{k}) - (\hat{j} 	imes \hat{i}) - (\hat{j} 	imes \hat{k})$$

Evaluating these cross products, we recall that $\hat{i} 	imes \hat{k} = -\hat{j}$, $\hat{j} 	imes \hat{i} = -\hat{k}$, and $\hat{j} 	imes \hat{k} = \hat{i}$. Substituting these in, we get:

$$\vec{n}_2 = 0 - \hat{j} - (-\hat{k}) - \hat{i} = -\hat{i} - \hat{j} + \hat{k}$$

### Phase 2: The Intersection Line

Now, we have our two normals: $\vec{n}_1 = \hat{k}$ and $\vec{n}_2 = -\hat{i} - \hat{j} + \hat{k}$. The line of intersection must be perpendicular to both of these normals.

Therefore, the direction vector $\vec{a}$ of our line is the cross product of $\vec{n}_1$ and $\vec{n}_2$:

$$\vec{a} = \vec{n}_1 	imes \vec{n}_2 = \hat{k} 	imes (-\hat{i} - \hat{j} + \hat{k})$$

Distributing the cross product, we find:

$$\vec{a} = -(\hat{k} 	imes \hat{i}) - (\hat{k} 	imes \hat{j}) + (\hat{k} 	imes \hat{k}) = -(\hat{j}) - (-\hat{i}) + 0 = \hat{i} - \hat{j}$$

This vector, $\vec{a} = \hat{i} - \hat{j}$, is the direction of the line where our two planes meet. It is the backbone of our geometry.

### Phase 3: The Final Angle

We are almost there. We need the angle $	heta$ between our vector $\vec{a} = \hat{i} - \hat{j}$ and the given vector $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$. We use the dot product formula:

$$\cos 	heta = \pm \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$

First, the dot product $\vec{a} \cdot \vec{b}$:

$$\vec{a} \cdot \vec{b} = (1)(1) + (-1)(-2) + (0)(2) = 1 + 2 = 3$$

Next, we calculate the magnitudes:

$$|\vec{a}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$

$$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

Plugging these into our cosine formula:

$$\cos 	heta = \pm \frac{3}{\sqrt{2} \cdot 3} = \pm \frac{1}{\sqrt{2}}$$

This gives us two possible values for $	heta$. If $\cos 	heta = \frac{1}{\sqrt{2}}$, then $	heta = \frac{\pi}{4}$. If $\cos 	heta = -\frac{1}{\sqrt{2}}$, then $	heta = \frac{3\pi}{4}$.

And there you have it! The angle between the line of intersection and the vector $\vec{b}$ is either **$\frac{\pi}{4}$** or **$\frac{3\pi}{4}$**. You have successfully navigated the intersection of two planes and mastered the vector algebra required to find the angle.

Visualized Solution

Visualizing the Planes

The Cross Product Logic

Setup for Normal $n_{1}$

Computing Normal $n_{1}$

Setup for Normal $n_{2}$

Computing Normal $n_{2}$

Setup for Vector $a$

Computing Vector $a$

Introducing Vector $b$

The Dot Product Formula

Setting up the Calculation

Computing the Values

Solving for $cos θ$

Final Angles

The Sigma Insight: Equation of a Plane

The Geometry of Intersecting Worlds

Phase 1

Phase 2

Phase 3

Similar Questions

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 length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - 2 \hat{k})$ is :

If the lines $r = (\hat{i} - \hat{j} + \hat{k}) + λ (3 \hat{j} - \hat{k})$ and $r = (α \hat{i} - \hat{j}) + μ (2 \hat{i} - 3 \hat{k})$ are co-planar, then distance of the plane containing these two lines from the point $(α, 0, 0)$ is :

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 largest value of $a$ , for which the perpendicular distance of the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + a \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - a \hat{k})$ from the point $(2, 1, 4)$ is $3$ , is ______.

Visualized Solution

Visualizing the Planes

The Cross Product Logic

Setup for Normal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​

Computing Normal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​

Setup for Normal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​

Computing Normal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​

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

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

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