Problem 99 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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 \(\vec{c}=\vec{a} \times \vec{b}\) then \(|\vec{a}|:|\vec{b}|:|\vec{c}| \quad[2002]\) (a) \(\sqrt{34}: \sqrt{45}: \sqrt{39}\) (b) \(\sqrt{34}: \sqrt{45}: 39\) (c) \(34: 39: 45\) (d) \(39: 35: 34\)

Step-by-Step Solution

Answer

The ratio is \( \sqrt{34}: \sqrt{45}: 39 \) (option b).

Step 1: Calculate the Magnitude of Vector \( \vec{a} \)

The vector \( \vec{a} \) is given as \( 3 \hat{i} - 5 \hat{j} \). To find its magnitude, we use the formula \( |\vec{a}| = \sqrt{a_1^2 + a_2^2} \). Here, \( a_1 = 3 \) and \( a_2 = -5 \), so:\[|\vec{a}| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}\]

Step 2: Calculate the Magnitude of Vector \( \vec{b} \)

The vector \( \vec{b} \) is given as \( 6 \hat{i} + 3 \hat{j} \). To find its magnitude, we also use the formula \( |\vec{b}| = \sqrt{b_1^2 + b_2^2} \). Here, \( b_1 = 6 \) and \( b_2 = 3 \), so:\[|\vec{b}| = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45}\]

Step 3: Calculate the Cross Product \( \vec{c} = \vec{a} \times \vec{b} \)

The cross product of two vectors \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \) gives:\[\vec{c} = (a_2b_3 - a_3b_2) \hat{i} - (a_1b_3 - a_3b_1) \hat{j} + (a_1b_2 - a_2b_1) \hat{k}\]For \( \vec{a} = 3 \hat{i} - 5 \hat{j} \), assume \( a_3 = 0 \), and for \( \vec{b} = 6 \hat{i} + 3 \hat{j} \), assume \( b_3 = 0 \). Thus:\[\vec{c} = (3*3 - (-5)*6) \hat{k} = (9 + 30) \hat{k} = 39 \hat{k}\]

Step 4: Calculate the Magnitude of Vector \( \vec{c} \)

Since \( \vec{c} = 39 \hat{k} \), its magnitude is:\[|\vec{c}| = \sqrt{39^2} = 39\]

Step 5: Determine the Ratio \(|\vec{a}|:|\vec{b}|:|\vec{c}|\)

Using the magnitudes calculated in the previous steps:\[|\vec{a}| = \sqrt{34}, \quad |\vec{b}| = \sqrt{45}, \quad |\vec{c}| = 39\]The ratio is \( \sqrt{34}: \sqrt{45}: 39 \).

Key Concepts

Vector Magnitude

The magnitude of a vector is like its "length". It gives you a single number that represents how long the vector is, ignoring its direction. To compute it in a two-dimensional space, you use the Pythagorean theorem. This is because each vector can be thought of as the hypotenuse of a right triangle. By summing the squares of its components, we find the square of its magnitude, as you have

• A vector \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} \) has a magnitude given by \( |\vec{a}| = \sqrt{a_1^2 + a_2^2} \).

In the original exercise, \( \vec{a} = 3 \hat{i} - 5 \hat{j} \), so the magnitude is \( \sqrt{3^2 + (-5)^2} = \sqrt{34} \). This process simplifies into carefully following the formula for any vector in 2D space.
Similarly, the vector \( \vec{b} = 6 \hat{i} + 3 \hat{j} \) has a magnitude \( \sqrt{45} \) calculated in the same fashion, illustrating how consistent and reliable the operation is.

Cross Product

The cross product of two vectors results in a third vector that is perpendicular to the plane formed by the original two vectors. This output vector's direction can be determined by the right-hand rule and its magnitude by the area of the parallelogram that the two vectors span in three-dimensional space. The cross product formula in 3D is

• \( \vec{a} \times \vec{b} = (a_2b_3 - a_3b_2) \hat{i} - (a_1b_3 - a_3b_1) \hat{j} + (a_1b_2 - a_2b_1) \hat{k} \).

In the example provided, since \( a_3 \) and \( b_3 \) are zero (as it is essentially a 2D problem enhanced to 3D for cross product computation), the cross product simplifies significantly:
\( \vec{c} = (3 \times 3 - (-5) \times 6) \hat{k} = (9 + 30) \hat{k} = 39 \hat{k} \).
This shows how even with zero in some dimensions, you can compute the cross product conveniently to find a meaningful result. The magnitude calculated as 39 shows the area represented.

Vector Operations

Vector operations, like addition, subtraction, and cross product, are foundational tools in vector mathematics. These operations allow us to manipulate vectors to solve problems involving both magnitude and direction. They extend to various physical and geometric applications, which makes them crucial in fields like physics and engineering.
For vector addition, each corresponding component is summed, so for vectors \( \vec{p} = p_1 \hat{i} + p_2 \hat{j} \) and \( \vec{q} = q_1 \hat{i} + q_2 \hat{j} \), the sum becomes:

• \( \vec{r} = (p_1 + q_1) \hat{i} + (p_2 + q_2) \hat{j} \).

Cross product, meanwhile, as shown earlier, yields a vector that can describe things like torque and rotational force.
Understanding these operations requires practice, and forms the basis for more complex studies in vector calculus, offering powerful insights into the behavior of physical systems. Applying these in problems like the JEE Main helps build a strong foundational skills set among students.