a = 3i - 5j and b = 6i + 3j are two vectors and c is a vector such that c = a x b then |a| : |b| : |

Q: Wait, why is the cross product only in the $z$-direction? Shouldn't it have $i$ and $j$ components too?

That is a fantastic question! Remember the geometric definition of the cross product: $\vec{a} \times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$. Since both $\vec{a}$ and $\vec{b}$ lie entirely within the $xy$-plane, any vector perpendicular to that plane must lie along the $z$-axis. Mathematically, when you expand the determinant, the $i$ and $j$ components vanish because the $z$-components of $\vec{a}$ and $\vec{b}$ are zero. You are essentially looking at the 'normal' to the plane!

Q: Can I use the formula $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta$ instead of the determinant?

You absolutely can! That is a perfectly valid approach. However, to use that, you would first need to find the angle $\theta$ between the vectors using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$. While that works, it often involves more steps and potential rounding errors with trigonometric functions. The determinant method is a direct, algebraic path that is usually faster and more robust for JEE problems.

Q: Why is the magnitude of $\vec{c}$ just 39 and not $\sqrt{39}$?

Great catch! Let's look at the result of our cross product: $\vec{c} = 39\hat{k}$. This vector has a magnitude of 39 because it is a vector pointing along the $z$-axis with a length of 39 units. The square root only appears when we calculate the magnitude of a vector with multiple components using the Pythagorean theorem. Since $\vec{c}$ has only one component, its magnitude is simply the absolute value of that component.

Question

$\vec{a} = 3\hat{i} - 5\hat{j}$ and $\vec{b} = 6\hat{i} + 3\hat{j}$ are two vectors and $\vec{c}$ is a vector such that $\vec{c} = \vec{a} 	imes \vec{b}$ then $|\vec{a}| : |\vec{b}| : |\vec{c}|$

Options:

(A) $\sqrt{34} : \sqrt{45} : \sqrt{39}$

(B) $\sqrt{34} : \sqrt{45} : 39$

(C) $34 : 39 : 45$

(D) $39 : 35 : 34$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are embarking on a journey to master the language of space itself: vectors. We are given two vectors, $\vec{a} = 3\hat{i} - 5\hat{j}$ and $\vec{b} = 6\hat{i} + 3\hat{j}$, resting comfortably in the two-dimensional $xy$-plane.

Our mission is to find the ratio of their magnitudes, including a mysterious third vector, $\vec{c}$, defined as the cross product of $\vec{a}$ and $\vec{b}$. This problem is not just about crunching numbers; it is about visualizing the hidden architecture of the coordinate system.

### Visualizing the Vectors

Imagine you are standing at the origin of a Cartesian coordinate system. Vector $\vec{a}$ is your first path, taking you 3 units to the right along the $x$-axis and 5 units down along the $y$-axis. It sits firmly in the fourth quadrant, a bold arrow pointing into the depths of the plane.

Now, look at vector $\vec{b}$. It takes you 6 units to the right and 3 units up, residing in the first quadrant. These two arrows define a plane, and our goal is to understand how they interact through the cross product.

### The Magnitude Calculation

Before we dive into the cross product, we must measure the lengths of these arrows. The magnitude of a vector is its length, and thanks to Pythagoras, we know that for any vector $\vec{v} = x\hat{i} + y\hat{j}$, the magnitude is $|\vec{v}| = \sqrt{x^2 + y^2}$.

For vector $\vec{a}$, we calculate:
$$|\vec{a}| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$$

For vector $\vec{b}$, we calculate:
$$|\vec{b}| = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45}$$

### The Cross Product: The Magic of the Determinant

Now, we come to the most exciting part: the cross product $\vec{c} = \vec{a} 	imes \vec{b}$. The cross product is a geometric operation that creates a new vector perpendicular to the plane containing the original two.

Since $\vec{a}$ and $\vec{b}$ are in the $xy$-plane, their cross product must point along the $z$-axis, in the direction of $\hat{k}$. To compute this, we use the determinant method:
$$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_x & a_y & a_z \ b_x & b_y & b_z \end{vmatrix}$$

Substituting our components, where $a_z = 0$ and $b_z = 0$, we get:
$$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -5 & 0 \ 6 & 3 & 0 \end{vmatrix}$$

Expanding this along the third column, we find:
$$\vec{c} = \hat{k} [3(3) - (-5)(6)] = \hat{k} [9 + 30] = 39\hat{k}$$

This is a beautiful result! The cross product has generated a vector pointing straight up along the $z$-axis with a magnitude of 39.

### Final Calculation

We have all our pieces: $|\vec{a}| = \sqrt{34}$, $|\vec{b}| = \sqrt{45}$, and $|\vec{c}| = 39$. The ratio we seek is:
$$|\vec{a}| : |\vec{b}| : |\vec{c}| = \sqrt{34} : \sqrt{45} : 39$$

This matches the required solution perfectly. You have navigated the geometry, mastered the algebra, and arrived at the **final answer** with precision.

Animated Solution for Mathematics - Vector Algebra: $a = 3 \hat{i} - 5 \hat{j}$ and $b = 6 \hat{i} + 3 \hat{j}$ are two vectors and $c$ is a vector such that $c = a \times b$ then $∣ a ∣ : ∣ b ∣ : ∣ c ∣$

Select Answer:

Visualized Solution

Visualizing the Vector Space

Plotting Vector $a = 3 \hat{i} - 5 \hat{j}$

Plotting Vector $b = 6 \hat{i} + 3 \hat{j}$

Calculating Magnitude $∣ a ∣$

Calculating Magnitude $∣ b ∣$

The Cross Product Concept

Computing $c = a \times b$

Magnitude of $c$

Finding the Ratio $∣ a ∣ : ∣ b ∣ : ∣ c ∣$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Visualizing the Vectors

The Magnitude Calculation

The Cross Product

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=3i^−5j^​ and b=6i^+3j^​ are two vectors and c is a vector such that c=a×b then ∣a∣:∣b∣:∣c∣

Select Answer:

Visualized Solution

Visualizing the Vector Space

Plotting Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=3i^−5j^​

Plotting Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=6i^+3j^​

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

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

The Cross Product Concept

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c=a×b

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

Finding the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a∣:∣b∣:∣c∣