Let a be a unit vector perpendicular to the vectors b=i-2j+3k and c=2i+3j-k, and makes an angle cos⁻

Q: Wait, why do we reject the positive root $\alpha = \sqrt{6}$? Isn't squaring both sides valid?

Squaring is a powerful tool, but it is also a dangerous one! It introduces extraneous solutions because it loses the information about the sign. Always check the sign of the original equation: $\frac{1}{2} = \frac{-\alpha}{\sqrt{3(2 + \alpha^2)}}$. Since the left side is positive, the right side must be positive, which forces $-\alpha$ to be positive, meaning $\alpha$ must be negative. Always respect the original equation before you square it!

Q: Why did we need to check the first angle condition with $\vec{d} = \hat{i} + \hat{j} + \hat{k}$? Couldn't we just pick either $\pm$?

Great question! The cross product gives us a line, but a line has two directions. The problem provides a specific constraint: the angle is $\cos^{-1}(-1/3)$. This constraint acts as a filter to select the correct orientation of $\hat{a}$ from the two possibilities ($\pm$) generated by the cross product. Without this check, you would be guessing blindly!

Q: Is there a faster way to find the unit vector without calculating the full cross product?

In this specific case, no. The cross product is the most robust tool to ensure orthogonality to two distinct vectors. While it feels like a lot of arithmetic, it is the only way to guarantee that $\hat{a}$ is perpendicular to the plane defined by $\vec{b}$ and $\vec{c}$. Trust the process—the calculation is straightforward and prevents geometric errors.

Question

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\vec{b}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{c}=2\hat{i}+3\hat{j}-\hat{k},$ and makes an angle $\cos^{-1}(-\frac{1}{3})$ with the vector $\hat{i}+\hat{j}+\hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i}+\alpha\hat{j}+\hat{k},$ then the value of $\alpha$ is:

Options:

(A) $\sqrt{6}$

(B) $-\sqrt{6}$

(C) $-\sqrt{3}$

(D) $\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the center of a vast, three-dimensional room. You have two vectors, $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{c} = 2\hat{i} + 3\hat{j} - \hat{k}$, stretching out from your feet. They define a plane, a flat surface slicing through the room.

Your mission is to find a unit vector, $\hat{a}$, that stands perfectly perpendicular to this plane. In the language of JEE physics and mathematics, when you need a vector perpendicular to two others, you summon the cross product.

The cross product $\vec{b} 	imes \vec{c}$ is the guardian of orthogonality. It provides a vector that obeys the right-hand rule, pointing straight out of the plane.

### The Calculation: Getting Our Hands Dirty

Let us perform the determinant expansion to find the normal vector. We set up our matrix:

$$
\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & 3 & -1 \end{vmatrix}
$$

Expanding this, we get $\hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4)$, which simplifies to $-7\hat{i} + 7\hat{j} + 7\hat{k}$.

Now, we calculate the magnitude to normalize this vector:

$$
|\vec{b} 	imes \vec{c}| = \sqrt{(-7)^2 + 7^2 + 7^2} = \sqrt{49 + 49 + 49} = 7\sqrt{3}
$$

Dividing our vector by this magnitude, we obtain the candidates $\hat{a} = \pm \frac{-\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.

### Selecting the Correct Vector

We are given that $\hat{a}$ makes an angle of $\cos^{-1}(-1/3)$ with $\vec{d} = \hat{i} + \hat{j} + \hat{k}$. Using the dot product formula $\hat{a} \cdot \frac{\vec{d}}{|\vec{d}|} = \cos(	heta)$, we test our candidates.

The candidate $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$ yields a dot product of $-1/3$, confirming it is our true vector.

### The Final Challenge: Solving for Alpha

We have a new vector $\vec{v} = \hat{i} + \alpha\hat{j} + \hat{k}$, and we know the angle between our confirmed $\hat{a}$ and $\vec{v}$ is $\frac{\pi}{3}$. We use the dot product formula:

$$
\cos\left(\frac{\pi}{3}ight) = \frac{\hat{a} \cdot \vec{v}}{|\hat{a}||\vec{v}|}
$$

Since $\cos(\frac{\pi}{3}) = 1/2$ and $|\hat{a}| = 1$, the equation becomes:

$$
\frac{1}{2} = \frac{\hat{a} \cdot \vec{v}}{|\vec{v}|} \Rightarrow \frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{2 + \alpha^2}}
$$

Squaring both sides to eliminate the radicals, we get:

$$
\frac{1}{4} = \frac{\alpha^2}{3(2 + \alpha^2)}
$$

Cross-multiplying gives $3(2 + \alpha^2) = 4\alpha^2$, which simplifies to $6 + 3\alpha^2 = 4\alpha^2$, leading us to $\alpha^2 = 6$.

This gives $\alpha = \pm\sqrt{6}$. However, looking back at the equation $\frac{1}{2} = \frac{-\alpha}{\sqrt{3(2 + \alpha^2)}}$, the left side is positive, so the right side must be positive.

This forces $-\alpha$ to be positive, meaning $\alpha$ must be negative. Thus, we reject the positive root and conclude that **$\alpha = -\sqrt{6}$**.

Select Answer:

Visualized Solution

Visualizing the Vectors

The Cross Product Concept

Calculating $b \times c$

Magnitude of the Cross Product

The Unit Vector Candidates

The First Angle Condition

Testing the Candidates

The Second Angle Condition

Setting up the Equation for $α$

Equating and Squaring

Solving the Quadratic Equation

Finalizing the Value of $α$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Calculation

Selecting the Correct Vector

The Final Challenge

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