Numerical Problems Based on Cross Product of two Vectors for Class 11 Physics

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Numerical Problems Based on Cross Product of two Vectors for Class 11 Physics

Numerical Problems on Vector Product (Cross Product)

Class 11 Physics · Motion in a Plane
SQ

Vector Cross Product Practice Set

Apply $\vec{A} \times \vec{B}$ determinants and area formulas
Ch 4 · Motion in a Plane
1
If $\vec{A} = \hat{i} + 3\hat{j} + 2\hat{k}$ and $\vec{B} = 3\hat{i} + \hat{j} + 2\hat{k}$, then find the vector product $\vec{A} \times \vec{B}$.
Answer $4\hat{i} + 4\hat{j} – 8\hat{k}$ 📝
Detailed Solution

The cross product $\vec{A} \times \vec{B}$ is calculated using the determinant method:

$$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 2 \\ 3 & 1 & 2 \end{vmatrix} $$

Expanding along the first row:

$$= \hat{i} \begin{vmatrix} 3 & 2 \\ 1 & 2 \end{vmatrix} – \hat{j} \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix}$$
$$= \hat{i}[(3)(2) – (2)(1)] – \hat{j}[(1)(2) – (2)(3)] + \hat{k}[(1)(1) – (3)(3)]$$
$$= \hat{i}(6 – 2) – \hat{j}(2 – 6) + \hat{k}(1 – 9)$$
$$= 4\hat{i} + 4\hat{j} – 8\hat{k}$$
2
Prove that the vectors $\vec{A} = 4\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = 12\hat{i} + 9\hat{j} + 3\hat{k}$ are parallel to each other.
Answer Proof completed 📝
Detailed Solution

Two vectors are parallel if their cross product is zero ($\vec{A} \times \vec{B} = 0$), or if their corresponding components are proportional.

Let’s check the ratio of their scalar components:

$$\frac{A_x}{B_x} = \frac{4}{12} = \frac{1}{3}$$
$$\frac{A_y}{B_y} = \frac{3}{9} = \frac{1}{3}$$
$$\frac{A_z}{B_z} = \frac{1}{3} = \frac{1}{3}$$

Since $\frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z}$, we can say that $\vec{B} = 3\vec{A}$. Because one is a scalar multiple of the other, the vectors are parallel.

3
If $\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{B} = 3\hat{i} + 2\hat{j} + 4\hat{k}$, then find the value of $(\vec{A} + \vec{B}) \times (\vec{A} – \vec{B})$.
Answer $-20\hat{i} + 10\hat{j} + 10\hat{k}$ 📝
Detailed Solution

Using the distributive property of the cross product:

$$(\vec{A} + \vec{B}) \times (\vec{A} – \vec{B}) = (\vec{A} \times \vec{A}) – (\vec{A} \times \vec{B}) + (\vec{B} \times \vec{A}) – (\vec{B} \times \vec{B})$$

We know that the cross product of a vector with itself is zero ($\vec{A} \times \vec{A} = 0$, $\vec{B} \times \vec{B} = 0$), and that $\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})$.

$$= 0 – (\vec{A} \times \vec{B}) – (\vec{A} \times \vec{B}) – 0 = -2(\vec{A} \times \vec{B})$$

Now, compute $\vec{A} \times \vec{B}$:

$$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 2 & 4 \end{vmatrix} $$
$$= \hat{i}(12 – 2) – \hat{j}(8 – 3) + \hat{k}(4 – 9) = 10\hat{i} – 5\hat{j} – 5\hat{k}$$

Multiply by $-2$:

$$-2(10\hat{i} – 5\hat{j} – 5\hat{k}) = -20\hat{i} + 10\hat{j} + 10\hat{k}$$
4
Find the value of $a$ for which the vectors $3\hat{i} + 3\hat{j} + 9\hat{k}$ and $\hat{i} + a\hat{j} + 3\hat{k}$ are parallel.
Answer $a = 1$ 📝
Detailed Solution

For two vectors to be parallel, the ratio of their corresponding components must be equal.

$$\frac{3}{1} = \frac{3}{a} = \frac{9}{3}$$

Taking the first two parts of the equation:

$$3 = \frac{3}{a} \implies 3a = 3 \implies a = 1$$
5
Find a unit vector perpendicular to the vectors $\vec{A} = 4\hat{i} – \hat{j} + 3\hat{k}$ and $\vec{B} = -2\hat{i} + \hat{j} – 2\hat{k}$.
Answer $\frac{1}{3}(-\hat{i} + 2\hat{j} + 2\hat{k})$ 📝
Detailed Solution

A vector perpendicular to both $\vec{A}$ and $\vec{B}$ is given by their cross product, $\vec{C} = \vec{A} \times \vec{B}$.

$$ \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} $$
$$= \hat{i}(2 – 3) – \hat{j}(-8 – (-6)) + \hat{k}(4 – 2)$$
$$= -\hat{i} – \hat{j}(-2) + 2\hat{k} = -\hat{i} + 2\hat{j} + 2\hat{k}$$

Next, find the magnitude of $\vec{C}$:

$$|\vec{C}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

The unit perpendicular vector $\hat{n}$ is $\frac{\vec{C}}{|\vec{C}|}$:

$$\hat{n} = \frac{1}{3}(-\hat{i} + 2\hat{j} + 2\hat{k})$$
6
Find the sine of the angle between the vectors $\vec{A} = 3\hat{i} – 4\hat{j} + 5\hat{k}$ and $\vec{B} = \hat{i} – \hat{j} + \hat{k}$.
Answer $1/5$ 📝
Detailed Solution

The sine of the angle $\theta$ is given by $\sin\theta = \frac{|\vec{A} \times \vec{B}|}{|\vec{A}||\vec{B}|}$.

1. Calculate $\vec{A} \times \vec{B}$:

$$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(-4 + 5) – \hat{j}(3 – 5) + \hat{k}(-3 + 4) = \hat{i} + 2\hat{j} + \hat{k}$$

2. Calculate the magnitudes:

$$|\vec{A} \times \vec{B}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$$
$$|\vec{A}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$$
$$|\vec{B}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$

3. Calculate $\sin\theta$:

$$\sin\theta = \frac{\sqrt{6}}{\sqrt{50} \times \sqrt{3}} = \frac{\sqrt{6}}{\sqrt{150}} = \frac{\sqrt{6}}{5\sqrt{6}} = \frac{1}{5}$$
7
Find a vector of magnitude 18 which is perpendicular to both the vectors $4\hat{i} – \hat{j} + 3\hat{k}$ and $-2\hat{i} + \hat{j} – 2\hat{k}$.
Answer $-6\hat{i} + 12\hat{j} + 12\hat{k}$ 📝
Detailed Solution

These are the identical vectors from Question 5. The unit vector $\hat{n}$ perpendicular to both is:

$$\hat{n} = \frac{1}{3}(-\hat{i} + 2\hat{j} + 2\hat{k})$$

To find a vector of magnitude 18 in this direction, multiply the unit vector by 18:

$$18\hat{n} = 18 \times \frac{1}{3}(-\hat{i} + 2\hat{j} + 2\hat{k}) = 6(-\hat{i} + 2\hat{j} + 2\hat{k})$$
$$= -6\hat{i} + 12\hat{j} + 12\hat{k}$$
8
Determine the area of the parallelogram whose adjacent sides are formed by the vectors $\vec{A} = \hat{i} – 3\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + \hat{j} + \hat{k}$.
Answer $4\sqrt{2}$ sq. units 📝
Detailed Solution

(Note: The textbook answer key likely contains a typo stating $\sqrt{2}$. The correct mathematical output is $4\sqrt{2}$).

The area of a parallelogram with adjacent sides $\vec{A}$ and $\vec{B}$ is given by $|\vec{A} \times \vec{B}|$.

$$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{vmatrix} $$
$$= \hat{i}(-3 – 1) – \hat{j}(1 – 1) + \hat{k}(1 – (-3))$$
$$= -4\hat{i} – 0\hat{j} + 4\hat{k}$$

Now, calculate the magnitude:

$$\text{Area} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\text{ square units}$$
9
Find the area of the triangle formed by points O, A and B such that $\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OB} = -3\hat{i} – 2\hat{j} + \hat{k}$.
Answer $3\sqrt{5}$ square units 📝
Detailed Solution

The area of a triangle formed by two position vectors originating from a common point is $\frac{1}{2}|\vec{OA} \times \vec{OB}|$.

$$ \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix} $$
$$= \hat{i}(2 – (-6)) – \hat{j}(1 – (-9)) + \hat{k}(-2 – (-6))$$
$$= 8\hat{i} – 10\hat{j} + 4\hat{k}$$

Now, find the magnitude of the cross product:

$$|\vec{OA} \times \vec{OB}| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}$$

Simplify $\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$.

$$\text{Area} = \frac{1}{2}(6\sqrt{5}) = 3\sqrt{5}\text{ square units}$$
10
Find with the help of vectors, the area of the triangle with vertices A(3, -1, 2), B(1, -1, -3) and C(4, -3, 1).
Answer $\sqrt{165}/2$ sq. units 📝
Detailed Solution

First, find the vectors for two sides of the triangle, e.g., $\vec{AB}$ and $\vec{AC}$.

$$\vec{AB} = (1-3)\hat{i} + (-1-(-1))\hat{j} + (-3-2)\hat{k} = -2\hat{i} + 0\hat{j} – 5\hat{k}$$
$$\vec{AC} = (4-3)\hat{i} + (-3-(-1))\hat{j} + (1-2)\hat{k} = 1\hat{i} – 2\hat{j} – 1\hat{k}$$

Calculate their cross product $\vec{AB} \times \vec{AC}$:

$$ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{vmatrix} $$
$$= \hat{i}(0 – 10) – \hat{j}(2 – (-5)) + \hat{k}(4 – 0) = -10\hat{i} – 7\hat{j} + 4\hat{k}$$

The area of the triangle is half the magnitude of this cross product:

$$|\vec{AB} \times \vec{AC}| = \sqrt{(-10)^2 + (-7)^2 + 4^2} = \sqrt{100 + 49 + 16} = \sqrt{165}$$
$$\text{Area} = \frac{\sqrt{165}}{2}\text{ sq. units}$$

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