Practice Questions on Vector Algebra (Hard)

Vector algebra is the branch of mathematics that deals with vector quantities that have both magnitude (size) and direction.

Question 1: Show that the points A(2\hat{i} - \hat{j} + \hat{k}), B(\hat{i} - 3\hat{j} - 5\hat{k}), C(3\hat{i} - 4\hat{j} - 4\hat{k}) are the vertices of a right-angled triangle.

Let,

=> \hat{a} = 2\hat{i} - \hat{j} + \hat{k}
=>\hat{b} = \hat{i} - 3\hat{j} - 5\hat{k}
=>\hat{c} = 3\hat{i} - 4\hat{j} - 4\hat{k}

The line segments are:

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

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

\vec{CA} = \vec{a} - \vec{c}
\vec{CA} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k})
\vec{CA} = -\hat{i} + 3\hat{j} + 5\hat{k}

The magnitude of the sides are,

|\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{41}
|\vec{BC}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{6}
|\vec{CA}| = \sqrt{(-1)^2 + (3)^2 + (5)^2} = \sqrt{35}

As we can see that |\vec{AB}|^2 =|\vec{BC}|^2 + |\vec{CA|^2}

=> Thus, ABC is a right-angled triangle.

Question 2: The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8\hat{i} - 6\hat{j} \text{ and } 3\hat{i} + 4\hat{j} - 12\hat{k}, is?

Diagonals of the parallogra is given by \vec{d_1} \text{ and } \vec{d_2} , then its area is calculated as: = \frac{1}{2} \left| \vec{d_1} \times \vec{d_2} \right|

Given,

\vec{d_1} = 8\hat{i} - 6\hat{j} + 0\hat{k},

\vec{d_2} = 3\hat{i} + 4\hat{j} - 12\hat{k}.

\vec{d_1} \times \vec{d_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\8 & -6 & 0 \\3 & 4 & -12\end{vmatrix}

= \hat{i} \begin{vmatrix} -6 & 0 \\ 4 & -12 \end{vmatrix} - \hat{j} \begin{vmatrix} 8 & 0 \\ 3 & -12 \end{vmatrix} + \hat{k} \begin{vmatrix} 8 & -6 \\ 3 & 4 \end{vmatrix}.

= \hat{i}((-6)(-12) - (0)(4)) - \hat{j}((8)(-12) - (0)(3)) + \hat{k}((8)(4) - (-6)(3)).

= \hat{i}(72) - \hat{j}(-96) + \hat{k}(50).

\vec{d_1} \times \vec{d_2} = 72\hat{i} + 96\hat{j} + 50\hat{k}.

\left| \vec{d_1} \times \vec{d_2} \right| = \sqrt{72^2 + 96^2 + 50^2}.

= \sqrt{16900} = 130

Area of parallelogram = \frac{1}{2} \left| \vec{d_1} \times \vec{d_2} \right| = \frac{1}{2} \times 130 = 65.

Question 3: If \vec{a}, \vec{b}, \vec{c} are vectors such that [\vec{a} \, \vec{b} \, \vec{c}] = 4, then \Big[(\vec{a} \times \vec{b}) \, (\vec{b} \times \vec{c}) \, (\vec{c} \times \vec{a})\Big] = ?

\Big[(\vec{a} \times \vec{b}) \, (\vec{b} \times \vec{c}) \, (\vec{c} \times \vec{a})\Big] = (\vec{a} \times \vec{b}) \cdot \Big((\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})\Big)

Simplify using vector triple product identity:

(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [\vec{c} \, \vec{a} \, \vec{b}] \vec{b} - [\vec{c} \, \vec{a} \, \vec{c}] \vec{c}

Substitude back:

(\vec{a} \times \vec{b}) \cdot \big((\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})\big) \\ = (\vec{a} \times \vec{b}) \big( [\vec{c} \, \vec{a} \, \vec{b}] \vec{b} - [\vec{c} \, \vec{a} \, \vec{c}] \vec{c}\big)

= \Big((\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c})\Big)^2

= \Big([\vec{a} \, \vec{b} \, \vec{c}]\Big)^2

= 4² = 16

Question 4: Let \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \, \vec{b} = 2\hat{i} - \hat{j} + \hat{k}, \, \vec{c} = x\hat{i} + (x + 1)\hat{j} + 2\hat{k}.

If the vector \vec{c} lies on the plane of \vec{a} and \vec{b} , then x equals?

Condition \vec{c} lies in the plane of \vec{a} \text{ and } \vec{b} \implies [\vec{a} \, \vec{b} \, \vec{c}] = 0.

[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix}1 & 2 & 3 \\2 & -1 & 1 \\x & x+1 & 2\end{vmatrix}

Expand the deteminant:

= 1 \cdot \begin{vmatrix}-1 & 1 \\x+1 & 2\end{vmatrix}- 2 \cdot \begin{vmatrix}2 & 1 \\x & 2\end{vmatrix}+ 3 \cdot \begin{vmatrix}2 & -1 \\x & x+1\end{vmatrix}

Compute each minor:

\begin{vmatrix}-1 & 1 \\x+1 & 2\end{vmatrix}= (-1)(2) - (1)(x+1) = -2 - x - 1 = -x - 3

\begin{vmatrix}2 & 1 \\x & 2\end{vmatrix}= (2)(2) - (1)(x) = 4 - x

\begin{vmatrix}2 & -1 \\x & x+1\end{vmatrix}= (2)(x+1) - (-1)(x) = 2x + 2 + x = 3x + 2

Subtitude back:

[\vec{a} \, \vec{b} \, \vec{c}] = 1(-x - 3) - 2(4 - x) + 3(3x + 2).

Simplify:

[\vec{a} \, \vec{b} \, \vec{c}] = -x - 3 - 8 + 2x + 9x + 6 = 10x - 5.

\text{Set } [\vec{a} \, \vec{b} \, \vec{c}] = 0:10x - 5 = 0 \implies x = \frac{1}{2}

x = 1/2

Question 5: For any vector \vec{b} , the value of \left|\vec{b} \times \hat{i}\right|^2 + \left|\vec{b} \times \hat{j}\right|^2 + \left|\vec{b} \times \hat{k}\right|^2 is equal to?

Let \vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}.

Compute: \vec{b} \times \hat{i}:

\vec{b} \times \hat{i} =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\b_x & b_y & b_z \\1 & 0 & 0\end{vmatrix}= b_z\hat{j} - b_y\hat{k}

Magnitude squared: \left|\vec{b} \times \hat{i}\right|^2 = b_z^2 + b_y^2

Compute: \vec{b} \times \hat{j}:

\vec{b} \times \hat{j} =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\b_x & b_y & b_z \\0 & 1 & 0\end{vmatrix}= b_x\hat{k} - b_z\hat{i}

Magnitude Squared: \left|\vec{b} \times \hat{k}\right|^2 = b_y^2 + b_x^2.

Sum of squares:

\left|\vec{b} \times \hat{i}\right|^2 + \left|\vec{b} \times \hat{j}\right|^2 + \left|\vec{b} \times \hat{k}\right|^2= (b_z^2 + b_y^2) + (b_x^2 + b_z^2) + (b_y^2 + b_x^2).

Simplify:

\left|\vec{b} \times \hat{i}\right|^2 + \left|\vec{b} \times \hat{j}\right|^2 + \left|\vec{b} \times \hat{k}\right|^2= 2(b_x^2 + b_y^2 + b_z^2).

Using |\vec{b}|^2 = b_x^2 + b_y^2 + b_z^2:

\left|\vec{b} \times \hat{i}\right|^2 + \left|\vec{b} \times \hat{j}\right|^2 + \left|\vec{b} \times \hat{k}\right|^2 = 2|\vec{b}|^2.

Question 6: Let \vec{a} = 9\hat{i} - 13\hat{j} + 25\hat{k}, \, \vec{b} = 3\hat{i} + 7\hat{j} - 13\hat{k}, \, \vec{c} = 17\hat{i} - 2\hat{j} + \hat{k} be three given vectors.

If \vec{r} is a vector such that \vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a} and \vec{r} \cdot (\vec{b} - \vec{c}) = 0 , then \frac{|593\vec{r} + 67\vec{a}|^2}{(593)^2} is equal to? [JEE Main 2024- April]

We are given:

\vec{a} = 9\hat{i} - 13\hat{j} + 25\hat{k},\\\vec{b} = 3\hat{i} + 7\hat{j} - 13\hat{k}, \\\vec{c} = 17\hat{i} - 2\hat{j} + \hat{k}

It is stated that:

\vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a}

\vec{r} - \vec{a} = (\vec{b} + \vec{c}) \times \vec{a} = 0

\vec{r} =(\vec{b} + \vec{c}) + \lambda \vec{a}

\vec{r} =(20\hat{i} + 5\hat{j}- 12\hat{k} ) + \lambda(9\hat{i} - 13\vec{r} + 25\hat{k})

= (20 + 9\lambda )\hat{i} + (5 -13\lambda)\hat{j} + (25\lambda - 12)\hat{k}

Now \vec{r} \cdot (\vec{b} - \vec{c}) = 0

\vec{r} =(-14\hat{i} + 9\hat{j}- 14\hat{k}) = 0

= -14 (20 + 9\lambda )\hat{i} + 9(5 -13\lambda)\hat{j} - 14 (25\lambda - 12)\hat{k} = 0

-593λ - 67 = 0

λ = -(67/593)

\therefore \vec{r} = (\vec{b} + \vec{c}) - \frac{67}{593} \vec{a}

\frac{|593\vec{r} + 67\vec{a}|^2}{|593|^2} = |\vec{b} + \vec{c}|^2 =|20\hat{i} + 5\hat{j}- 12\hat{k}|^2

= 569

Question 7: Let \lambda \in \mathbb{R}, \, \vec{a} = \lambda \hat{i} + 2\hat{j} - 3\hat{k}, \, \vec{b} = \hat{i} - \lambda \hat{j} + 2\hat{k}.

If ((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}, then |\vec{a} + \vec{b}| \times |\vec{a} - \vec{b}|^2 is equal to? [JEE Main 2023- January]

((\vec{a} \times \vec{b}) + \vec{b} \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b})

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

= 8 (\vec{a} \times \vec{b})

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

= \hat{i}(4 - 3\lambda) - \hat{j}(2\lambda + 3) + \hat{k}(\lambda^2 - 2)

\lambda = 1

|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 = 140

Question 8: Let \vec{a} = 2\hat{i} - \hat{j} + 5\hat{k} \text{ and } \vec{b} = \alpha \hat{i} + \beta \hat{j} + 2\hat{k}.

Then \text{If } ((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}, \text{ then } |\vec{b} \times 2\hat{j}| is equal to? [JEE Main 2022- July]

Given: \vec{a} = 2\hat{i} - \hat{j} + 5\hat{k} \text{ and } \vec{b} = \alpha \hat{i} + \beta \hat{j} + 2\hat{k}.

Also, ((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}.

\Rightarrow ((\vec{a} \cdot \hat{i}) \vec{b} - (\vec{b} \cdot \hat{i}) \vec{a}) \cdot \hat{k} = \frac{23}{2}.

\Rightarrow (2 \cdot \vec{b} - \alpha \cdot \vec{a}) \cdot \hat{k} = \frac{23}{2}.

\Rightarrow 2 \cdot 2 - 5\alpha = \frac{23}{2} \Rightarrow \alpha = -\frac{3}{2}.

Now, |\vec{b} \times 2\hat{j}| = |(\alpha \hat{i} + \beta \hat{j} + 2\hat{k}) \times 2\hat{j}|.

= |2 \cdot 2\hat{k} + 0 - 4\hat{i}|

= \sqrt{4\alpha^2 + 16}.

= \sqrt{4 \left(-\frac{3}{2}\right)^2 + 16} = 5.

= 5

Practice Questions

Question 1: The area ( in square units) of the paralleogram whose diagonals are along the vectors 7\hat{i} + 3\hat{j} \text{ and } 4\hat{i} + 6\hat{j} - 8\hat{k} , is?

Question 2: If \vec{x}, \vec{y}, \vec{z} are vectors such that [\vec{x} \, \vec{y} \, \vec{z}] = 5 , then find \Big[(\vec{x} \times \vec{y}) \, (\vec{y} \times \vec{z}) \, (\vec{z} \times \vec{x})\Big] .

Question 3: Let \vec{p} = \hat{i} + 4\hat{j} + 3\hat{k}, \, \vec{q} = 3\hat{i} - \hat{j} + 2\hat{k}, \, \vec{r} = x\hat{i} + (x - 2)\hat{j} + \hat{k}.

If the vector \vec{r} lies in the plane of \vec{p} \text{ and } \vec{q}, find x.

Question 4: For any vector \vec{v} , calculate the value of \left|\vec{v} \times \hat{i}\right|^2 + \left|\vec{v} \times \hat{j}\right|^2 + \left|\vec{v} \times \hat{k}\right|^2.

Question 5: The values of a, for which the points P, Q, R with position vectors 3\hat{i} + \hat{j} - 2\hat{k}, \, -\hat{i} + 4\hat{j} + \hat{k}, \, \text{and } a\hat{i} + \hat{j} + 2\hat{k},

respectively, are the vertices of a right-angled triangle with ∠PQR=π/2 are?

Question 6: Let \vec{p} = 2\hat{i} + \hat{j} - \hat{k}, \, \vec{q} = \hat{i} - 2\hat{j} + 3\hat{k}, \text{ and } \vec{r} = x\hat{i} + (x+1)\hat{j} - 2\hat{k}.

If the vector \vec{r} lies in the plane of \vec{q} , then x equals?

Question 7: If \vec{u}| = 5, \, |\vec{v}| = 3 and the angle between \vec{u} and \vec{v} is π/4 then \Big|\vec{u} \times \vec{v}\Big|^2 is equal to?

Question 8: If the vectors \vec{c}, \vec{a} = x\hat{i} + y\hat{j} + z\hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} are such that

\vec{a}, \vec{c}, \text{ and } \vec{b} form a right-handed system, then \vec{c} is?

Answer Key

1. 33.96
2. 25
3. 3/2
4. 2|\vec{v}_x|^2 + 2|\vec{v}_y|^2 + 2|\vec{v}_z|^2
5. -(5/2)
6. 1/2
7. 112.5
8. \vec{c} = (-z)\hat{i} + z\hat{j} + (x - y)\hat{k}