QJEEM23Ph8fe5: JEE Main 2023 Physics - Class 11 JEEM23Ph8fe5 Solution: When vector is subtracted from vector , it gives a vector equal …

Given that when vector $\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ is subtracted from vector $\overrightarrow{\mathrm{B}}$ , it gives a vector equal to $2 \hat{j}$ .

We can write this as:

$\vec{B} - \vec{A} = 2 \hat{j}$

Now, let's express the vector $\overrightarrow{\mathrm{B}}$ in terms of its components:

$\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

Subtract vector $\vec{A}$ from vector $\vec{B}$ :

$(B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) - (2 \hat{i}+3 \hat{j}+2 \hat{k}) = 2 \hat{j}$

Comparing the components, we get:

$B_x - 2 = 0 \\B_y - 3 = 2 \\B_z - 2 = 0$

Solving these equations, we find the components of vector $\vec{B}$ :

$B_x = 2 \\B_y = 5 \\B_z = 2$

Now, we can find the magnitude of vector $\vec{B}$ :

$|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{2^2 + 5^2 + 2^2} = \sqrt{4 + 25 + 4} = \sqrt{33}$

Therefore, the magnitude of vector $\overrightarrow{\mathrm{B}}$ is $\sqrt{33}$