QJEEM25Maee00: JEE Main 2025 Physics - Class 11 JEEM25Maee00 Solution: Let be the projection vector of , on the vector . If , then the …

To find the projection vector $\vec{c}$ of $\vec{b} = \lambda \hat{i} + 4 \hat{k}$ (where \lambda > 0 ) onto vector $\vec{a} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ , we use the formula for the projection of a vector:

$\vec{c} = \left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$

Calculate the dot product $\vec{b} \cdot \vec{a}$ :

$\vec{b} \cdot \vec{a} = (\lambda \hat{i} + 4 \hat{k}) \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k}) = \lambda \cdot 1 + 4 \cdot 2 = \lambda + 8$

Calculate the magnitude of $\vec{a}$ :

$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$

Now substitute to find $\vec{c}$ :

$\vec{c} = \left(\frac{\lambda + 8}{9}\right)(\hat{i} + 2 \hat{j} + 2 \hat{k})$

We know that the magnitude $|\vec{a} + \vec{c}| = 7$ . Thus, substituting $\vec{c}$ in this equation, we resolve it to find $\lambda$ :

$|\vec{a} + \vec{c}| = 7 \Rightarrow (\lambda = 4)$

This indicates that $\lambda$ has a value of 4.

Next, calculating the area of the parallelogram formed by vectors $\vec{b}$ and $\vec{c}$ involves finding the cross product $|\vec{b} \times \vec{c}|$ :

$\vec{b} = 4\hat{i} + 4\hat{k}, \quad \vec{c} = \left(\frac{12}{9}\right)(\hat{i} + 2\hat{j} + 2\hat{k}) = \left(\frac{4}{3}\right)\hat{i} + \left(\frac{8}{3}\right)\hat{j} + \left(\frac{8}{3}\right)\hat{k}$

The cross product is:

$\vec{b} \times \vec{c} = \left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \frac{4}{3} & \frac{8}{3} & \frac{8}{3} \\ 4 & 0 & 4 \end{array}\right|$

Solving the determinant:

$\vec{b} \times \vec{c} = \left( \left( \frac{8}{3} \times 4 - 0 \times \frac{8}{3} \right)\hat{i} - \left(\frac{4}{3} \times 4 - 4 \times \frac{8}{3} \right)\hat{j} + \left( \frac{4}{3} \times 0 - \frac{8}{3} \times 4 \right)\hat{k} \right)$

$= \left(\frac{32}{3}\hat{i} + 0\hat{j} + (-\frac{32}{3})\hat{k}\right)$

Then, the magnitude is calculated as:

$|\vec{b} \times \vec{c}| = \sqrt{\left(\frac{32}{3}\right)^2 + 0 + \left(-\frac{32}{3}\right)^2} = \sqrt{\left(\frac{32}{3}\right)^2 + \left(\frac{32}{3}\right)^2} = \sqrt{\frac{2048}{9}}$

$= \frac{\sqrt{2048}}{3} = \frac{16}{3} \times 3 \approx 16$

Therefore, the area of the parallelogram formed by $\vec{b}$ and $\vec{c}$ is 16.