Let p = 2i + 3j + k and q = i + 2j + k be two vectors. If a vector r = (αi + βj + γk) is perpendicul

Q: Wait, why do we use (p + q) and (p - q) instead of just p and q?

That is a brilliant observation! While you could use p and q directly, the problem explicitly states r is perpendicular to (p + q) and (p - q). Mathematically, the plane spanned by (p + q) and (p - q) is identical to the plane spanned by p and q. Using the sum and difference often simplifies the arithmetic, making the cross product calculation much cleaner.

Q: Do I really need to worry about the plus-minus sign when finding r?

In this specific case, because we are asked for the sum of the absolute values |alpha| + |beta| + |gamma|, the sign of the components will disappear. However, in JEE Advanced, always keep the plus-minus sign in your working until the very end to ensure you don't miss a potential second solution or violate a constraint.

Q: Is there a faster way to find the cross product without the determinant?

The determinant method is the gold standard for reliability. While you can use distributive properties of the cross product, the determinant is less prone to silly calculation errors. In the heat of the exam, stick to the method that guarantees accuracy.

Q: Why is the magnitude of r equal to sqrt(3)?

The magnitude is a constraint given to fix the scale of the vector. Without it, there would be infinitely many vectors perpendicular to the plane. By setting |r| = sqrt(3), we lock the vector into a specific length, allowing us to find unique values for alpha, beta, and gamma.

Q: What if I get a different vector direction?

If your cross product direction is the negative of mine, don't panic! A vector perpendicular to a plane can point in two opposite directions. Since the final question asks for the sum of absolute values, both directions will yield the exact same result.

Question

Let $\vec{p} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + 2\hat{j} + \hat{k}$ be two vectors. If a vector $\vec{r} = (\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k})$ is perpendicular to each of the vectors $(\vec{p} + \vec{q})$ and $(\vec{p} - \vec{q})$, and $|\vec{r}| = \sqrt{3}$, then $|\alpha| + |\beta| + |\gamma|$ is equal to

Sigma Percentile · Accepted Answer

### The Geometry of Orthogonality: A Journey into Vector Space

Welcome, fellow traveler of the JEE landscape! Today, we are not just solving a problem; we are exploring the elegant architecture of 3D space.

We are given two vectors, $\vec{p} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{q} = \hat{i} + 2\hat{j} + \hat{k}$. Our mission is to find a mysterious vector $\vec{r} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ that stands perfectly perpendicular to two specific directions: the sum $(\vec{p} + \vec{q})$ and the difference $(\vec{p} - \vec{q})$.

### Phase 1: Simplifying the Landscape

Before we dive into the heavy machinery of cross products, let us simplify our terrain. We need to find the vectors that define our plane of interest.

First, the sum: 
$$\vec{p} + \vec{q} = (2+1)\hat{i} + (3+2)\hat{j} + (1+1)\hat{k} = 3\hat{i} + 5\hat{j} + 2\hat{k}$$

Next, the difference: 
$$\vec{p} - \vec{q} = (2-1)\hat{i} + (3-2)\hat{j} + (1-1)\hat{k} = \hat{i} + \hat{j}$$

Notice how the difference vector simplifies beautifully to $\hat{i} + \hat{j}$. This is the beauty of vector algebra—often, the most complex-looking expressions hide simple, elegant truths.

### Phase 2: The Power of the Cross Product

Now, we face the core of the problem. If a vector $\vec{r}$ is perpendicular to two other vectors, it must be parallel to their cross product.

Let us define a vector $\vec{v}$ as the cross product of our sum and difference vectors: 
$$\vec{v} = (\vec{p} + \vec{q}) 	imes (\vec{p} - \vec{q})$$

To compute this, we invoke the determinant, the trusty tool of every vector analyst: 
$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 5 & 2 \ 1 & 1 & 0 \end{vmatrix}$$

Expanding this, we get: 
$$\vec{v} = \hat{i}(5 \cdot 0 - 2 \cdot 1) - \hat{j}(3 \cdot 0 - 2 \cdot 1) + \hat{k}(3 \cdot 1 - 5 \cdot 1) = -2\hat{i} + 2\hat{j} - 2\hat{k}$$

This vector $\vec{v}$ is our compass. It points in the direction of $\vec{r}$.

### Phase 3: Scaling to Perfection

We know $\vec{r}$ is parallel to $\vec{v}$, so $\vec{r} = k\vec{v}$. But what is $k$? We are given the constraint $|\vec{r}| = \sqrt{3}$.

First, let us find the magnitude of our direction vector $\vec{v}$: 
$$|\vec{v}| = \sqrt{(-2)^2 + 2^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$$

Now, we normalize $\vec{v}$ to find the unit vector $\hat{r}$: 
$$\hat{r} = \pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{-2\hat{i} + 2\hat{j} - 2\hat{k}}{2\sqrt{3}} = \pm \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} - \hat{k})$$

Finally, we scale this unit vector by the required magnitude $\sqrt{3}$: 
$$\vec{r} = \sqrt{3} \cdot \left[ \pm \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} - \hat{k}) ight] = \pm(-\hat{i} + \hat{j} - \hat{k})$$

This gives us our components: $\alpha = \mp 1$, $\beta = \pm 1$, and $\gamma = \mp 1$.

### The Final Celebration

$$|\alpha| + |\beta| + |\gamma| = |-1| + |1| + |-1| = 1 + 1 + 1 = \mathbf{3}$$

And there we have it! Through the systematic application of vector properties, we have navigated the space and arrived at the solution. Remember, in physics and math, the path is just as important as the destination.

Enter Numerical Value:

Visualized Solution

Defining Vectors $p$ and $q$

Calculating the Sum $(p + q)$

Calculating the Difference $(p - q)$

The Orthogonality Condition

Setting up the Cross Product

Evaluating the Determinant

Finding the Unit Vector

Determining Vector $r$

Final Calculation

The Sigma Insight: Vector (Cross) Product

The Geometry of Orthogonality

Phase 1

Phase 2

Phase 3

The Final Celebration

Similar Questions

Let $a = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - 2 \hat{k}$ be two vectors. If a vector perpendicular to both the vectors $a + b$ and $a - b$ has the magnitude 12 then one such vector is

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . If $c$ is a vector such that $a \cdot c = ∣ c ∣, ∣ c - a ∣ = 22$ and the angle between $(a \times b)$ and $c$ is $3 0^{\circ}$ , then $∣ (a \times b) \times c ∣ =$

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Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}, b = \hat{i} + \hat{j}$ and $c = a \times b$ . Let $d$ be a vector such that $∣ d - a ∣ = 11, ∣ c \times d ∣ = 3$ and the angle between $c$ and $d$ is $\frac{π}{4}$ . Then $a \cdot d$ is equal to

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Let $α = 3 \hat{i} + \hat{j}$ and $β = 2 \hat{i} - \hat{j} + 3 \hat{k}$ . If $β = β_{1} - β_{2}$ , where $β_{1}$ is parallel to $α$ and $β_{2}$ is perpendicular to $α$ , then $β_{1} \times β_{2}$ is equal to

Let $a = \hat{i} + 2 \hat{j} - \hat{k}, b = \hat{i} - \hat{j}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ and $r \cdot b = 0$ , then $r \cdot a$ is equal to

Let $a = \hat{j} - \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ . Then the vector $b$ satisfying $a \times b + c = 0$ and $a \cdot b = 3$

Enter Numerical Value:

Visualized Solution

Defining Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q​

Calculating the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (p​+q​)

Calculating the Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (p​−q​)