Physics Vectors & Equilibrium PMDC Conceptual Practice
PMDC Verified Question 34 of 50
Given vectors \(\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{B} = 4\hat{i} + 2\hat{j} - 2\hat{k}\), find a vector \(\vec{x}\) that is parallel to \(\vec{A}\) but has a magnitude equal to that of \(\vec{B}\).
A
\(\frac{2\sqrt{21}}{7}(2\hat{i} + 3\hat{j} - \hat{k})\)
B
\(\sqrt{\frac{7}{12}}(4\hat{i} + 2\hat{j} - 2\hat{k})\)
C
\(\sqrt{\frac{7}{12}}(2\hat{i} + 3\hat{j} + \hat{k})\)
D
\(\sqrt{\frac{3}{5}}(\hat{i} + 2\hat{j} - \hat{k})\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \(\frac{2\sqrt{21}}{7}(2\hat{i} + 3\hat{j} - \hat{k})\)
Concept:
Scaling a unit vector to construct a vector with a specific direction and magnitude.

Formula:
$$\vec{x} = |\vec{B}| \hat{A} = |\vec{B}| \frac{\vec{A}}{|\vec{A}|}$$

Solution:
First, calculate the scalar magnitude of both vectors:
$$|\vec{B}| = \sqrt{4^2 + 2^2 + (-2)^2} = \sqrt{16 + 4 + 4} = \sqrt{24}$$
$$|\vec{A}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$$
Now substitute these values into our equation for \(\vec{x}\):
$$\vec{x} = \sqrt{24} \frac{2\hat{i} + 3\hat{j} - \hat{k}}{\sqrt{14}} = \sqrt{\frac{24}{14}}(2\hat{i} + 3\hat{j} - \hat{k})$$
Simplify the radical fraction:
$$\sqrt{\frac{24}{14}} = \sqrt{\frac{12}{7}} = \sqrt{\frac{12 \times 7}{7^2}} = \frac{\sqrt{84}}{7} = \frac{2\sqrt{21}}{7}$$
Therefore, the constructed vector is:
$$\vec{x} = \frac{2\sqrt{21}}{7}(2\hat{i} + 3\hat{j} - \hat{k})$$

Why other options are incorrect:
  • B, C, and D use incorrect magnitude ratios or incorrect vector components that do not point in the direction of \(\vec{A}\).

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