Let a be a non-zero vector parallel to the line of intersection of the two planes described by i + j

Q: Why do we need the scalar $\lambda$? Can't we just use the cross product result as $\vec{a}$?

Great question! The cross product $\vec{n}_1 \times \vec{n}_2$ only gives the *direction* of the line. The vector $\vec{a}$ might be longer or shorter than this cross product. The scalar $\lambda$ acts as a scaling factor to ensure $\vec{a}$ satisfies the specific condition $\vec{a} \cdot \vec{b} = 6$. Without it, you are assuming the magnitude of the intersection vector is fixed, which isn't stated.

Q: Is there a faster way to find the angle $\theta$ without calculating the cross product magnitude first?

Absolutely! You can use the dot product formula $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. Since you know the dot product is $6$ and you can calculate the magnitudes of $\vec{a}$ and $\vec{b}$ once you have found $\lambda$, you can find $\cos \theta$ directly. This is often safer than calculating the cross product magnitude, which involves more algebra.

Q: I keep getting a sign error when calculating the cross product. How do I avoid this?

Sign errors are the most common trap in JEE vector problems! Always write out the determinant explicitly: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix}$. When expanding along the first row, remember the alternating signs: positive for $\hat{i}$, negative for $\hat{j}$, and positive for $\hat{k}$. Writing that negative sign for the $\hat{j}$ component before you even start calculating the minor will save you every time.

Question

Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i} + \hat{j}, \hat{i} + \hat{k}$ and $\hat{i} - \hat{j}, \hat{j} - \hat{k}$. If $	heta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{a} \cdot \vec{b} = 6$, then the ordered pair $(	heta, |\vec{a} 	imes \vec{b}|)$ is equal to

Options:

(A) $(\frac{\pi}{3}, 3\sqrt{6})$

(B) $(\frac{\pi}{4}, 3\sqrt{6})$

(C) $(\frac{\pi}{3}, 6)$

(D) $(\frac{\pi}{4}, 6)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the vector $\vec{a}$ that lies on the line of intersection of two planes, we must first determine the normal vectors to these planes. The first plane is spanned by vectors $\vec{u} = \hat{i} + \hat{j}$ and $\vec{v} = \hat{i} + \hat{k}$.

The normal vector $\vec{n}_1$ is calculated via the cross product:
$$\vec{n}_1 = (\hat{i} + \hat{j}) 	imes (\hat{i} + \hat{k}) = \hat{i} - \hat{j} - \hat{k}$$

Similarly, the second plane is spanned by $\vec{w} = \hat{i} - \hat{j}$ and $\vec{z} = \hat{j} - \hat{k}$. Its normal vector $\vec{n}_2$ is:
$$\vec{n}_2 = (\hat{i} - \hat{j}) 	imes (\hat{j} - \hat{k}) = \hat{i} + \hat{j} + \hat{k}$$

### Finding the Direction of Intersection

The line of intersection is perpendicular to both normal vectors $\vec{n}_1$ and $\vec{n}_2$. Therefore, the direction of this line is given by the cross product $\vec{n}_1 	imes \vec{n}_2$:
$$\vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -1 \ 1 & 1 & 1 \end{vmatrix} = 0\hat{i} - 2\hat{j} + 2\hat{k}$$

Since the vector $\vec{a}$ is parallel to this line, we can express it as:
$$\vec{a} = \lambda(-2\hat{j} + 2\hat{k})$$

### Solving for the Vector $\vec{a}$

We are given the condition $\vec{a} \cdot \vec{b} = 6$, where $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$. Substituting our expression for $\vec{a}$ into this dot product:
$$\lambda(-2\hat{j} + 2\hat{k}) \cdot (2\hat{i} - 2\hat{j} + \hat{k}) = 6$$
$$\lambda(0 + 4 + 2) = 6 \Rightarrow 6\lambda = 6 \Rightarrow \lambda = 1$$

Thus, the vector is determined to be $\vec{a} = -2\hat{j} + 2\hat{k}$.

### Final Calculation

First, we calculate the magnitudes of the vectors:
$$|\vec{a}| = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$
$$|\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$$

Next, we find the angle $	heta$ between $\vec{a}$ and $\vec{b}$:
$$\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{6}{(2\sqrt{2})(3)} = \frac{1}{\sqrt{2}}$$
This implies $	heta = \frac{\pi}{4}$.

Finally, the magnitude of the cross product is:
$$|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}| \sin 	heta = (2\sqrt{2})(3) \left( \frac{1}{\sqrt{2}} ight) = 6$$

The final result is **6**.

Select Answer:

Visualized Solution

Visualizing the Intersection of Planes

Finding Normal to Plane 1: $n_{1}$

Computing $n_{1}$

Finding Normal to Plane 2: $n_{2}$

Computing $n_{2}$

Direction of Intersection Line

Calculating the Cross Product

Using the Dot Product Condition

Solving for $λ$

Final Vector $a$ and Magnitudes

Calculating the Angle $θ$

Magnitude of Cross Product

Final Ordered Pair

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Finding the Direction of Intersection

Solving for the Vector $a$

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Intersection of Planes

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

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

Finding Normal to Plane 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n2​

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

Direction of Intersection Line

Calculating the Cross Product

Using the Dot Product Condition

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

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

Calculating the Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ