Question 44: Motion in 2D & Vectors

Question

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 $$2 \hat{j}$$. Then the magnitude of vector $$\overrightarrow{\mathrm{B}}$$ will be :

Accepted Answer

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 $$\sqrt{33}$$

Answer

3

Answer

$$\sqrt{33}$$

Answer

$$\sqrt6$$

Answer

$$\sqrt5$$

Motion in 2D & Vectors

When vector $\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k} is subtracted from vector \overrightarrow{\mathrm{B}}, it gives a vector equal to 2 \hat{j}. Then the magnitude of vector \overrightarrow{\mathrm{B}}$ will be :

Choose the correct answer: