Vector Algebra Practice Questions (Medium)

A vector is a quantity that has both magnitude and direction, and vector algebra provides the tools to perform calculations and solve problems involving vectors.

Question 1: Find the dot product of the vectors:\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = 4\hat{i} + \hat{j} + 5\hat{k} .

Then calculate the angle θ between the two vectors.

The dot product of two vectors \vec{a} and \vec{b} is given by:

\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z

Subtitute the components of \vec{a} and \vec{b} :

\vec{a}\cdot \vec{b} = (1)(4) + (2)(1) + (3)(5) = 21

The angle θ between two vectors is given by:\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}​

Magnitude of the Vectors

|\vec{a}| = \sqrt{1^2+ 2^2+(3)^2} = \sqrt{14}

|\vec{b}| = \sqrt{4^2+ (1)^2+(5)^2} = \sqrt{42}

\cos\theta = \frac{21}{\sqrt{14}\cdot\sqrt{42}}

After Factorizing the denominator

Cos θ = \frac{\sqrt3}{2}

\theta = \cos^{-1} (\frac{\sqrt3}{2}) = 30\degree or \frac{\pi}{6} radions.

Question 2: The vectors \vec{a} = 3\hat{i} - 4\hat{j} + \hat{k},\space \vec{b} = \hat{i} + 2\hat{j} - \hat{k} form two adjacent sides of a parallelogram. Find:

1. \vec{a} \times \vec{b},
2. The area of the parallelogram.

The formula for the cross product of two vectors is: \vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\a_x & a_y & a_z \\b_x & b_y & b_z\end{vmatrix}

Subtitue the values: \vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & -4 & 1 \\1 & 2 & -1\end{vmatrix}

Expand the determinant:

\vec{a} \times \vec{b} = \hat{i} \begin{vmatrix}-4 & 1 \\2 & -1\end{vmatrix} \space-\hat{j} \begin{vmatrix}3 & 1 \\1 & -1\end{vmatrix} \space+\hat{k} \begin{vmatrix}3 & -4 \\1 & 2\end{vmatrix} \space

Compute each minor determinant:

1. For \hat{i} = \begin{vmatrix}-4 & 1 \\2 & -1\end{vmatrix} \space = (-4)(-1) - (2)(1) = 4 -2 = 2

2. For \hat{j} = \begin{vmatrix}3 & 1 \\1 & -1\end{vmatrix} \space = (3)(−1) − (1)(1) = −3 − 1 = −4

3. For \hat{k} = \begin{vmatrix}3 & -4 \\1 & 2\end{vmatrix} \space =(3)(2) − (1)(−4) = 6 + 4 = 10

\vec{a} \times \vec{b} = 2\hat{i} -( -4)\hat{j} + 10\hat{k}\\\\ \\\vec{a} \times \vec{b} = 2\hat{i} + 4\hat{j} + 10\hat{k}

Area of the Parallelogram

The area of the parallelogram is given by the magnitude of the cross product: \text{Area} = |\vec{a} \times \vec{b}|

The magnitude of \vec{a} \times \vec{b} = 2\hat{i} + 4\hat{j} + 10\hat{k} is: |\vec{a} \times \vec{b}| = \sqrt{2^2+4^2+10^2} = 2\sqrt{30}

Question 3: Given:

\vec{a} = 2\hat{i} + \hat{j} - \hat{k}, \quad \vec{b} = \hat{i} - \hat{j} + \hat{k}, \quad \vec{c} = 3\hat{i} - \hat{j} + 2\hat{k}

Find (\vec{a} \times \vec{b}) \cdot \vec{c}

Compute:\vec{a} \times \vec{b}

The formula for the cross product of two vectors is:\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\a_x & a_y & a_z \\b_x & b_y & b_z\end{vmatrix}

Subtitue the values:\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\2 & 1 & -1 \\1 & -1 & -\end{vmatrix}

Expand the determinant:

\vec{a} \times \vec{b} = \hat{i} \begin{vmatrix}1 & 1 \\-1 & 1\end{vmatrix} \space-\hat{j} \begin{vmatrix}2 & -1 \\1 & 1\end{vmatrix} \space+\hat{k} \begin{vmatrix}2 & 1\\1 & -1\end{vmatrix} \space

Compute each minor determinant:

1. For \hat{i} = = (-4)(-1) - (2)(1) = 4 -2 = 2

2. For \hat{j} = = (3)(−1) − (1)(1) = −3 − 1 = −4

3. For \hat{k} = =(3)(2) − (1)(−4) = 6 + 4 = 10

\vec{a} \times \vec{b} = 0\hat{i} -3\hat{j} - 3\hat{k}\\\\ \\\vec{a} \times \vec{b} = - 3\hat{j} - 3\hat{k}

Compute (\vec{a} \times \vec{b}) \ \cdot \vec{c}

The dot product of two vectors is:

(\vec{a} \times \vec{b}) \cdot \vec{c} = (-3\hat{j} - 3\hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k})

Expand using the dot product formula:

(\vec{a} \times \vec{b}) \cdot \vec{c} = (−3)(−1) + (−3)(2)

Perform the calculations:

(\vec{a} \times \vec{b}) \cdot \vec{c} = 3 - 6 = -3

Question 4: Find the vector equation of a line passing through the points ((1, 2, 3) and (4, 5, 6). Express the line in the form \vec{r} = \vec{a} + \lambda \vec{b} , where λ is a scalar.

Given: Points (1, 2, 3) and (4, 5, 6)

The vector equation of a line is:

\vec{r} = \vec{a} + \lambda \vec{b},

where \vec{a} is a position vector of one point, and \vec{b} is the direction vector.

Position vector of the first point:

\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}

Direction vector:

\vec{b} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}

Equation of the line:

\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} + 3\hat{j} + 3\hat{k})

Simplify:

\vec{r} = (1 + 3\lambda)\hat{i} + (2 + 3\lambda)\hat{j} + (3 + 3\lambda)\hat{k}

Question 5: Find the shortest distance between the skew lines:

• \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda (2\hat{i} - \hat{j} + \hat{k})
• \vec{r} = 3\hat{i} - \hat{j} + 4\hat{k} + \mu (\hat{i} + \hat{j} + 2\hat{k})

Given:

\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda (2\hat{i} - \hat{j} + \hat{k})

\vec{r} = 3\hat{i} - \hat{j} + 4\hat{k} + \mu (\hat{i} + \hat{j} + 2\hat{k})

Direction vectors of the lines:

\vec{d_1} = 2\hat{i} - \hat{j} + \hat{k}, \quad \vec{d_2} = \hat{i} + \hat{j} + 2\hat{k}

Vector joining a point on Line 1 and Line 2:

\vec{n} = (3 - 1)\hat{i} + (-1 - 2)\hat{j} + (4 - 3)\hat{k} = 2\hat{i} - 3\hat{j} + \hat{k}.

Shortest distance formula:

\frac{|\vec{n} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}

Cross Product \vec{d_1} \times \vec{d_2} ​​:

\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 1 & 2 \end{vmatrix}

\vec{d_1} \times \vec{d_2} = -3\hat{i} - 3\hat{j} + 3\hat{k}

Magnitude of Cross Product:

|\vec{d_1} \times \vec{d_2}| = \sqrt{(-3)^2 + (-3)^2 + (3)^2} = 3\sqrt{3}.

Dot Product \vec{n} \cdot (\vec{d_1} \times \vec{d_2}) :

\vec{n} \cdot (\vec{d_1} \times \vec{d_2}) = (2)(-3) + (-3)(-3) + (1)(3) = -6 + 9 + 3 = 6.

Shortest Distance: d = \frac{|6|}{3\sqrt{3}} = \frac{2}{\sqrt{3}}

Question 6: Find a vector parallel to the plane 2x + 3y − z = 52 that is perpendicular to the vector \vec{A} = \hat{i} + \hat{j} + \hat{k}.

Given:
Plane 2x + 3y − z = 52, vector \vec{A} = \hat{i} + \hat{j} + \hat{k}

A vector parallel to the plane is perpendicular to the plane's normal vector:

\vec{N} = 2\hat{i} + 3\hat{j} - \hat{k}

Let \vec{P} = a\hat{i} + b\hat{j} + c\hat{k}. For \vec{P} to be parallel to the plane:

\vec{P} \cdot \vec{N}= 0

Substitute:

(a\hat{i} + b\hat{j} + c\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0

2a + 3b − c = 0.

Assume a = 1, b = 1, and solve for c:

2(1) + 3(1) − c = 0  ⟹  c = 5

Parallel vector:

\vec{P} = \hat{i} + \hat{j} + 5\hat{k}

Question 7: Can a vector have direction angles α = 30^∘, β = 45^∘, and γ = 135^∘? If yes, find the direction cosines of the vector.

The direction cosines are given by: l = cos ⁡α, m = cos ⁡β, n =cos ⁡γ

Substitute the angles: l = cos⁡30^∘ = √3/2, m = cos⁡45^∘ = √2/2, n = cos⁡135^∘ = −(√2/2)

The condition for direction cosines is: l² + m² + n² = 1

Subtitute l = √3/2, m = √2/2, n = −(√2/2):

(\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{2}}{2})^2 + (-\frac{\sqrt{2}}{2})^2 = 1

Add them together:

3/2 + 2/4 + 2/4 = 7/4

The sum of the squares of the direction cosines is:

l² + m² + n² ≠ 1

Thus, a vector cannot have direction angles α = 30^∘, β = 45^∘, and γ = 135^∘.

Question 8: A vector makes angles π/3​ with the x-axis and π/4​ with the y-axis. Find the angle it makes with the z-axis.

We know that if l, m and n are the direction cosines and  α, β and γ are the direction angles then,

l = cos ⁡α = cos π/3 = cos⁡ 60° = 1/2,

m = cos ⁡β = cos⁡ π/4 = Cos 45° = √2/2

We need to find γ, the angle with the z-axis.

Solve for n²

Using the property l² + m² + n² = 1, substitute l and m:

\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + n^2 = 1.

1/4 + 2/4 + n² = 1

3/4 + n² = 1

n² = 1 - 3/4

n² = 1/4

n = \pm\sqrt{\frac{1}{4}} = \pm\frac{1}{2}

Since n = cos γ, we only take only positive values of n, assuming the vector points ibn the positive z-direction: n = 1/2

cos γ = n = 1/2

γ = cos^-1 (1/2) = π/3

The vector makes an angle of: γ = π/3 (60 degrees) with the z-axis

Unsolved Practice Questions

Question 1: Find the dot product of the vectors: \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}, \quad \vec{b} = 4\hat{i} - \hat{j} + 5\hat{k}, Then calculate the angle θ between the two vectors.

Question 2: The vectors: \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = 2\hat{i} - \hat{j} + \hat{k} represent two adjacent sides of a parallelogram. Find:

• The cross product \vec{a} \times \vec{b}.
• The area of the parallelogram.

Question 3: Given: \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}, \quad \vec{b} = \hat{i} + \hat{j} - 2\hat{k}, \quad \vec{c} = 3\hat{i} - 2\hat{j} + \hat{k},

find the scalar triple product (\vec{a} \times \vec{b}) \cdot \vec{c}

Question 4: Find the vector equation of a line passing through the points: P(2, −1, 3) and Q(4, 3, −2). Express the equation in the form:

\vec{r} = \vec{a} + \lambda \vec{b}.

Question 5: Find the shortest distance between the skew lines: \vec{r_1} = 2\hat{i} + \hat{j} - \hat{k} + \lambda(3\hat{i} - 2\hat{j} + \hat{k}) and \vec{r_2} = -\hat{i} + 2\hat{j} + 3\hat{k} + \mu(2\hat{i} + \hat{j} - \hat{k}).

Question 6: Find a vector parallel to the plane x − 2y + 3z = 7 that is perpendicular to the vector \vec{A} = 2\hat{i} - \hat{j} + \hat{k}

Question 7: Can a vector have direction angles α = 60^∘, β = 60^∘, γ = 60^∘? If yes, find the direction cosines.

Question 8: A vector makes an angle π/6 with the x-axis and π/4​ with the y-axis. Find the angle it makes with the z-axis.

Answer Key

1. Dot product = 0, angle between them = 90^∘.
2. Cross product = 5\hat{i} + 5\hat{j} - 5\hat{k} , Area of parallelogram = 5 √3​.
3. −14
4. \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 4\hat{j} - 5\hat{k})
5. 2√3
6. \hat{i} + 5\hat{j}+3\hat{k}
7. Cannot have these direction angles.
8. π/3​ or 60^∘.