Let a = i + 5j + αk, b = i + 3j + βk and c = -i + 2j - 3k be three vectors such that, |b × c| = 5√3

Q: Wait, why do we square the magnitude of the cross product? Can't we just solve it with the square root?

You absolutely could, but in the heat of a JEE Advanced exam, efficiency and accuracy are paramount. Squaring both sides, $|\vec{b} \times \vec{c}|^2 = 75$, eliminates the radical immediately. It transforms a potentially confusing radical equation into a clean, standard quadratic equation. It is a tactical choice to minimize calculation errors.

Q: Why did we get two values for $\beta$? Does that mean the problem has two answers?

Great observation! The problem asks for the *greatest* amongst the values of $|\vec{a}|^2$. This implies that there are multiple valid configurations for these vectors. Both values of $\beta$ are mathematically valid, leading to two distinct possible vectors $\vec{a}$. We simply calculate the magnitude for both and pick the winner.

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

While you could use the distributive property of the cross product, the determinant method is the gold standard for 3D vectors. It is systematic, less prone to sign errors, and keeps your work organized. Stick to the determinant—it is the most reliable tool in your arsenal.

Question

Let $\vec{a} = \hat{i} + 5\hat{j} + \alpha\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} + \beta\hat{k}$ and $\vec{c} = -\hat{i} + 2\hat{j} - 3\hat{k}$ be three vectors such that, $|\vec{b} 	imes \vec{c}| = 5\sqrt{3}$ and $\vec{a}$ is perpendicular to $\vec{b}$. Then the greatest amongst the values of $|\vec{a}|^2$ is .

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Variables: A Vector Odyssey

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a 3D puzzle. When you look at vectors $\vec{a} = \hat{i} + 5\hat{j} + \alpha\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} + \beta\hat{k}$, and $\vec{c} = -\hat{i} + 2\hat{j} - 3\hat{k}$, do not see them as mere lists of numbers. See them as arrows in space, defined by their orientation and their length.

We have two mysterious scalars, $\alpha$ and $\beta$, hiding in the shadows. Our mission is to unmask them using the two clues provided: orthogonality and the magnitude of a cross product.

### Phase 1: The Orthogonality Anchor

Let us start with the most elegant property of vectors: orthogonality. When we say $\vec{a} \perp \vec{b}$, we are saying that the angle between them is exactly $90^\circ$. In the language of linear algebra, this is a gift. It tells us that their dot product must vanish into nothingness.

We write: 
$$\vec{a} \cdot \vec{b} = 0$$

Substituting the components, we perform the dot product: $(1)(1) + (5)(3) + (\alpha)(\beta) = 0$. This simplifies beautifully to $1 + 15 + \alpha\beta = 0$, or simply $\alpha\beta = -16$. Hold onto this relationship; it is the bridge that will connect our two variables later.

### Phase 2: The Cross Product Challenge

Now, we turn to the second clue: $|\vec{b} 	imes \vec{c}| = 5\sqrt{3}$. This is where many students stumble, not because the math is hard, but because the algebra can get messy. We need to compute the cross product $\vec{b} 	imes \vec{c}$ by setting up our determinant:

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

Expand this with care, remembering the alternating signs for the determinant expansion. For the $\hat{i}$ component, we have $(3)(-3) - (\beta)(2) = -9 - 2\beta$. For the $\hat{j}$ component, we have $-((1)(-3) - (\beta)(-1)) = -(-3 + \beta) = 3 - \beta$. Finally, for the $\hat{k}$ component, we have $(1)(2) - (3)(-1) = 2 + 3 = 5$.

So, our cross product vector is $(-9-2\beta)\hat{i} + (3-\beta)\hat{j} + 5\hat{k}$.

### Phase 3: The Quadratic Bridge

We are given that the magnitude of this vector is $5\sqrt{3}$. Instead of dealing with the square root, we square both sides: $|\vec{b} 	imes \vec{c}|^2 = (5\sqrt{3})^2 = 75$.

Now, we sum the squares of the components:

$$(-9-2\beta)^2 + (3-\beta)^2 + 5^2 = 75$$

Expanding these terms is a test of your algebraic discipline. We get $(81 + 36\beta + 4\beta^2) + (9 - 6\beta + \beta^2) + 25 = 75$. Combining like terms, we arrive at $5\beta^2 + 30\beta + 115 = 75$.

Subtracting $75$ from both sides, we get $5\beta^2 + 30\beta + 40 = 0$. Dividing by $5$, we find the elegant quadratic: $\beta^2 + 6\beta + 8 = 0$. Factoring this is a joy: $(\beta + 4)(\beta + 2) = 0$. Thus, $\beta$ can be $-4$ or $-2$.

### Phase 4: The Final Calculation

We are almost there. We have two scenarios for $\beta$, and consequently, two scenarios for $\alpha$ (since $\alpha\beta = -16$).

**Case 1:** If $\beta = -4$, then $\alpha = 4$. The magnitude squared of $\vec{a}$ is:
$$|\vec{a}|^2 = 1^2 + 5^2 + 4^2 = 1 + 25 + 16 = 42$$

**Case 2:** If $\beta = -2$, then $\alpha = 8$. The magnitude squared of $\vec{a}$ is:
$$|\vec{a}|^2 = 1^2 + 5^2 + 8^2 = 1 + 25 + 64 = 90$$

The problem asks for the greatest value. Comparing $42$ and $90$, the answer is clear. You have navigated the complexity, handled the algebra, and arrived at the truth. The maximum value is **90**.

Enter Numerical Value:

Visualized Solution

Analyze Given Vectors

Apply Orthogonality Condition

Calculate Dot Product

Relation Between $α$ and $β$

Setup Cross Product $b \times c$

Expand the Determinant

Magnitude Condition

Expand and Simplify

Solve for $β$

Case 1: $β = - 4$

Case 2: $β = - 2$

Final Conclusion

The Sigma Insight: Vector (Cross) Product

The Geometry of Hidden Variables

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Consider three vectors $a, b, c$ . Let $∣ a ∣ = 2, ∣ b ∣ = 3$ and $a = b \times c$ . If $α \in [0, \frac{π}{3}]$ is the angle between the vectors $b$ and $c$ , then the minimum value of $27∣ c - a ∣^{2}$ is equal to:

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 $∣ (a \times b) \times c ∣ =$

Let $p = 2 \hat{i} + 3 \hat{j} + \hat{k}$ and $q = \hat{i} + 2 \hat{j} + \hat{k}$ be two vectors. If a vector $r = (α \hat{i} + β \hat{j} + γ \hat{k})$ is perpendicular to each of the vectors $(p + q)$ and $(p - q)$ , and $∣ r ∣ = 3$ , then $∣ α ∣ + ∣ β ∣ + ∣ γ ∣$ is equal to

Let a vector $a$ has a magnitude 9. Let a vector $b$ be such that for every $(x, y) \in R \times R - {(0, 0)}$ , the vector $(x a + y b)$ is perpendicular to the vector $(6 y a - 18 x b)$ . Then the value of $∣ a \times b ∣$ is equal to:

Let $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ , $b = \hat{i} - 2 \hat{j} - 2 \hat{k}$ and $c = - \hat{i} + 4 \hat{j} + 3 \hat{k}$ . If $d$ is a vector perpendicular to both $b$ and $c$ , and $a \cdot d = 18$ , then $∣ a \times d ∣^{2}$ is equal to

Let $a = 2 \hat{i} - \hat{j} + 3 \hat{k}$ , $b = 3 \hat{i} - 5 \hat{j} + \hat{k}$ and $c$ be a vector such that $a \times c = c \times b$ and $(a + c) \cdot (b + c) = 168$ Then the maximum value of $∣ c ∣^{2}$ is:

Let $a, b, c$ be three vectors such that $∣ a ∣ = 31, 4∣ b ∣ = ∣ c ∣ = 2$ and $2 (a \times b) = 3 (c \times a)$ . If the angle between $b$ and $c$ is $2 π /3$ , then $(\frac{a \times c}{a \cdot b})^{2}$ is equal to

Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = \hat{i} - \hat{j} + 2 \hat{k}$ and $c = 5 \hat{i} - 3 \hat{j} + 3 \hat{k}$ be there vectors. If $r$ is a vector such that, $r \times b = c \times b$ and $r \cdot a = 0$ . Then $25∣ r ∣^{2}$ is equal to