Vectors Algebra
55 Questions
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Q51
DPT
Dot Product
MCQ
23 Jul 2026
Question 9 in the document is:
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Correct Answer:
Q52
DPT
Dot Product
MCQ
23 Jul 2026
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If vector $\vec{C}$ is added to vector $\vec{D}$, the result is a third vector that is perpendicular to $\vec{D}$ and has a magnitude equal to $3D$. What is the ratio of the magnitude of $\vec{C}$ to that of $\vec{D}$?
**Concept:**
Vector addition, vector components, and the Pythagorean theorem for right-angled triangles formed by perpendicular vectors.
**Options:**
A) 1.8
B) 2.2
C) 3.2
D) 1.3
**Correct Answer:**
C
**Explanation:**
Let the sum of vectors $\vec{C}$ and $\vec{D}$ be $\vec{R} = \vec{C} + \vec{D}$.
Rearranging this gives $\vec{C} = \vec{R} - \vec{D}$.
Since $\vec{R}$ is perpendicular to $\vec{D}$, the vectors $\vec{R}$ and $\vec{D}$ form the legs of a right-angled triangle, and vector $\vec{C}$ represents the hypotenuse.
Using the Pythagorean theorem:
$C^2 = R^2 + D^2$
Given that $R = 3D$:
$C^2 = (3D)^2 + D^2$
$C^2 = 9D^2 + D^2$
$C^2 = 10D^2$
$C = \sqrt{10} D$
The ratio of the magnitude of $\vec{C}$ to that of $\vec{D}$ is:
$\frac{C}{D} = \sqrt{10} \approx 3.16 \approx 3.2$
If vector $\vec{C}$ is added to vector $\vec{D}$, the result is a third vector that is perpendicular to $\vec{D}$ and has a magnitude equal to $3D$. What is the ratio of the magnitude of $\vec{C}$ to that of $\vec{D}$?
**Concept:**
Vector addition, vector components, and the Pythagorean theorem for right-angled triangles formed by perpendicular vectors.
**Options:**
A) 1.8
B) 2.2
C) 3.2
D) 1.3
**Correct Answer:**
C
**Explanation:**
Let the sum of vectors $\vec{C}$ and $\vec{D}$ be $\vec{R} = \vec{C} + \vec{D}$.
Rearranging this gives $\vec{C} = \vec{R} - \vec{D}$.
Since $\vec{R}$ is perpendicular to $\vec{D}$, the vectors $\vec{R}$ and $\vec{D}$ form the legs of a right-angled triangle, and vector $\vec{C}$ represents the hypotenuse.
Using the Pythagorean theorem:
$C^2 = R^2 + D^2$
Given that $R = 3D$:
$C^2 = (3D)^2 + D^2$
$C^2 = 9D^2 + D^2$
$C^2 = 10D^2$
$C = \sqrt{10} D$
The ratio of the magnitude of $\vec{C}$ to that of $\vec{D}$ is:
$\frac{C}{D} = \sqrt{10} \approx 3.16 \approx 3.2$
Correct Answer:
Q53
DPT
Resultant
MCQ
23 Jul 2026
Concept: System of linear vector equations and solving for unknown vectors by elimination.
Given that $\vec{A} + 2\vec{B} = x_1\hat{i} + y_1\hat{j}$ and $2\vec{A} - \vec{B} = x_2\hat{i} + y_2\hat{j}$, what is $\vec{A}$?
A.
$\vec{A} = \frac{1}{5}(x_1 + 2x_2)\hat{i} + \frac{1}{5}(y_1 + 2y_2)\hat{j}$
B.
$\vec{A} = \frac{1}{5}(x_1 + 2x_2)\hat{i} - \frac{1}{5}(y_1 + 2y_2)\hat{j}$
C.
$\vec{A} = \frac{1}{5}(x_1 + 4x_2)\hat{i} + \frac{1}{5}(y_1 + 2y_2)\hat{j}$
D.
$\vec{A} = \frac{1}{5}(x_1 + 4x_2)\hat{i} + \frac{1}{5}(y_1 + 4y_2)\hat{j}$
Q54
DPT
Resultant
MCQ
23 Jul 2026
Concept: Vector subtraction, isosceles triangles formed by rotating a vector of fixed magnitude, and the small-angle approximation $\sin\left(\frac{\Delta\theta}{2}\right) \approx \frac{\Delta\theta}{2}$.
A vector $\vec{A}$ is rotated by a small angle $\Delta\theta$ radians ($\Delta\theta \ll 1$) to get a new vector $\vec{B}$. In that case $\vert{}\vec{B} - \vec{A}\vert{}$ is
A.
$0$
B.
$\vert{}\vec{A}\vert{} \left(1 - \frac{\Delta\theta^2}{2}\right)$
C.
$\vert{}\vec{A}\vert{} \Delta\theta$
D.
$\vert{}\vec{B}\vert{} \Delta\theta - \vert{}\vec{A}\vert{}$
Q55
DPT
Resultant
MCQ
23 Jul 2026
Concept: Vector subtraction, isosceles triangles formed by rotating a vector of fixed magnitude, and the small-angle approximation $\sin\left(\frac{\Delta\theta}{2}\right) \approx \frac{\Delta\theta}{2}$.
A vector $\vec{A}$ is rotated by a small angle $\Delta\theta$ radians ($\Delta\theta \ll 1$) to get a new vector $\vec{B}$. In that case $\vert{}\vec{B} - \vec{A}\vert{}$ is
A.
$0$
B.
$\vert{}\vec{A}\vert{} \left(1 - \frac{\Delta\theta^2}{2}\right)$
C.
$\vert{}\vec{A}\vert{} \Delta\theta$
D.
$\vert{}\vec{B}\vert{} \Delta\theta - \vert{}\vec{A}\vert{}$