Question

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

• A. $\sqrt{34} : \sqrt{45} : 39$
• B. $34 : 45 : 39$
• C. $34 : 39 : 45$
• D. $39 : 35 : 34$

Hint:

When crossing two vectors that only have $\hat{i}$ and $\hat{j}$ components, you don't need a full $3 \times 3$ determinant! Just use the shortcut: $(x_1 y_2 - x_2 y_1)\hat{k}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two 2D vectors in the xy-plane. We must compute their cross product to find a third vector $\bar{c}$, and then calculate the ratio of the magnitudes of all three vectors.

Step 2: Key Formula or Approach:
The magnitude of a vector $\bar{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\bar{v}| = \sqrt{x^2 + y^2 + z^2}$.
The cross product of two vectors in the xy-plane ($\bar{a} = a_x\hat{i} + a_y\hat{j}$ and $\bar{b} = b_x\hat{i} + b_y\hat{j}$) results in a vector purely in the z-direction: $\bar{a}\times\bar{b} = (a_x b_y - a_y b_x)\hat{k}$.

Step 3: Detailed Explanation:
First, find the magnitudes of $\bar{a}$ and $\bar{b}$:
$$|\bar{a}| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$$ $$|\bar{b}| = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45}$$ Next, calculate the cross product $\bar{c} = \bar{a} \times \bar{b}$:
$$\bar{c} = (3\hat{i} - 5\hat{j}) \times (6\hat{i} + 3\hat{j})$$ $$\bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{vmatrix}$$ $$\bar{c} = \hat{k}( (3)(3) - (-5)(6) )$$ $$\bar{c} = \hat{k}( 9 + 30 ) = 39\hat{k}$$ Now, find the magnitude of $\bar{c}$:
$$|\bar{c}| = \sqrt{0^2 + 0^2 + 39^2} = 39$$ The required ratio $|\bar{a}|:|\bar{b}|:|\bar{c}|$ is therefore $\sqrt{34} : \sqrt{45} : 39$.
(Note: Option A in the raw exam data is often misprinted without the square root symbols, but mathematically it corresponds to the magnitudes derived here.)

Step 4: Final Answer:
The ratio is $\sqrt{34} : \sqrt{45} : 39$, corresponding to option (A).