Let a = αi + 3j + k, b = 3i - βj + 4k and c = i + 2j - 2k where α, β ∈ R, be three vectors. If the p

Q: Why do we divide by the magnitude of $\vec{c}$ in the projection formula?

Great question! The dot product $\vec{a} \cdot \vec{c}$ is defined as $|\vec{a}||\vec{c}|\cos\theta$. The projection of $\vec{a}$ onto $\vec{c}$ is defined as the component of $\vec{a}$ along the direction of $\vec{c}$, which is $|\vec{a}|\cos\theta$. If you look at the dot product formula, you see that to get $|\vec{a}|\cos\theta$, you must divide the dot product by $|\vec{c}|$. It is all about isolating the scalar component!

Q: I got $\beta = 1$ from the $\hat{i}$ component, but what if the $\hat{k}$ component gave a different value?

In a consistent mathematical system, the components must align. If you find that the $\hat{i}$ component gives one value for $\beta$ and the $\hat{k}$ component gives another, it is a massive red flag that a sign error occurred during the determinant expansion. Always use the other components to verify your result—it is the best way to catch silly mistakes before they cost you marks.

Q: The options provided don't include 7. Did I make a mistake?

Trust your math! In competitive exams, sometimes there are errors in the question paper. If your derivation is solid, and you have verified your steps, stand by your result. In this case, the math leads us to 7, and that is the absolute truth of the problem.

Question

Let $\vec{a} = \alpha\hat{i} + 3\hat{j} + \hat{k}, \vec{b} = 3\hat{i} - \beta\hat{j} + 4\hat{k}$ and $\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}$ where $\alpha, \beta \in \mathbb{R}$, be three vectors. If the projection of $\vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\vec{b} 	imes \vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$, then the value of $\alpha + \beta$ equal to:

Options:

(A) 3

(B) 4

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system with three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$. These are not just lists of numbers; they are physical entities with direction and magnitude.

Our mission is to find the values of $\alpha$ and $\beta$ that define these vectors. This is the language of physics and geometry.

### The Shadow of a Vector

We begin with the projection of $\vec{a}$ onto $\vec{c}$. Think of this as shining a light perpendicular to $\vec{c}$ and looking at the shadow $\vec{a}$ casts. The formula is:

$$\mathrm{Proj}_{\vec{c}}\vec{a} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}$$

First, we calculate the magnitude of $\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}$. Using the Pythagorean theorem in 3D:

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

Next, we compute the dot product $\vec{a} \cdot \vec{c} = (\alpha)(1) + (3)(2) + (1)(-2) = \alpha + 6 - 2 = \alpha + 4$. Equating this to the given projection of $\frac{10}{3}$:

$$\frac{\alpha + 4}{3} = \frac{10}{3}$$

The denominators cancel out, leaving us with $\alpha + 4 = 10$, which yields **$\alpha = 6$**. We have successfully pinned down the first unknown.

### The Power of the Cross Product

Now, we turn to the cross product $\vec{b} 	imes \vec{c}$. The cross product creates a vector that is mutually perpendicular to both $\vec{b}$ and $\vec{c}$. We set up the determinant:

$$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -\beta & 4 \ 1 & 2 & -2 \end{vmatrix}$$

Expanding this determinant, we get:

$$\hat{i}[(-\beta)(-2) - (4)(2)] - \hat{j}[(3)(-2) - (4)(1)] + \hat{k}[(3)(2) - (-\beta)(1)]$$

Simplifying each component, we obtain:

$$(2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$$

The problem states this equals $-6\hat{i} + 10\hat{j} + 7\hat{k}$. By comparing the $\hat{i}$ components:

$$2\beta - 8 = -6 \Rightarrow 2\beta = 2 \Rightarrow \beta = 1$$

We verify this with the $\hat{k}$ component: $6 + \beta = 7$, which confirms **$\beta = 1$**. The consistency is beautiful.

### The Synthesis

We have found $\alpha = 6$ and $\beta = 1$. The final step is simple arithmetic:

$$\alpha + \beta = 6 + 1 = 7$$

The final result is **$7$**. You have navigated the projection and the cross product with precision, which is the essence of JEE Advanced preparation.

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Visualized Solution

Given Vectors

Projection of $a$ on $c$

Magnitude of $c$

Dot Product $a \cdot c$

Solving for $α$

The Cross Product Condition

Determinant for $b \times c$

Expanding the Determinant

Solving for $β$

Final Sum $α + β$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Shadow of a Vector

The Power of the Cross Product

The Synthesis

Similar Questions

Let $a = α \hat{i} + \hat{j} - \hat{k}$ and $b = 2 \hat{i} + \hat{j} - α \hat{k}, α > 0$ . If the projection of $a \times b$ on the vector $- \hat{i} + 2 \hat{j} - 2 \hat{k}$ is 30, then $α$ is equal to

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Let $a = \hat{i} - 3 \hat{j} + 7 \hat{k}$ , $b = 2 \hat{i} - \hat{j} + \hat{k}$ and $c$ be a vector such that $(a + 2 b) \times c = 3 (c \times a)$ . If $a \cdot c = 130$ , then $b \cdot c$ is equal to _______

Let $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $b = 7 \hat{i} + \hat{j} - 6 \hat{k}$ . If $r \times a = r \times b$ , $r \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = - 3$ , then $r \cdot (2 \hat{i} - 3 \hat{j} + \hat{k})$ is equal to

Let $a$ be a vector which is perpendicular to the vector $3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$ . If $a \times (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}$ , then the projection of the vector $a$ on the vector $2 \hat{i} + 2 \hat{j} + \hat{k}$ is

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

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Let $a = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $b = 2 \hat{i} - 3 \hat{j} + 5 \hat{k}$ . If $r \times a = b \times r$ , $r \cdot (α \hat{i} + 2 \hat{j} + \hat{k}) = 3$ and $r \cdot (2 \hat{i} + 5 \hat{j} - α \hat{k}) = - 1$ , $α \in R$ , then the value of $α + ∣ r ∣^{2}$ is equal to :

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Visualized Solution

Given Vectors

Projection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅c

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