Solved Examples

Important Theorems and Concepts Recap

1. Triangle Law of Vector Addition

If two vectors are represented by two sides of a triangle taken in order, then their sum is represented by the third side taken from the starting point of the first vector to the ending point of the second vector. If $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$, then $$\vec{AB} + \vec{BC} = \vec{AC}.$$ This law gives the geometric meaning of vector addition.

2. Section Formula

If a point $R$ divides the line segment joining points with position vectors $\vec{a}$ and $\vec{b}$:

• Internally in the ratio $m:n$, $$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$$
• Externally in the ratio $m:n$, $$\vec{r} = \frac{m\vec{b} - n\vec{a}}{m-n}$$

3. Scalar (Dot) Product

• Definition: $$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$$
• Angle between vectors: $$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$
• Scalar projection of $\vec{a}$ on $\vec{b}$: $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$

4. Vector (Cross) Product

• Definition: $$\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \, \hat{n}$$
• Area of triangle with adjacent sides $\vec{a}, \vec{b}$: $$\frac{1}{2}|\vec{a} \times \vec{b}|$$
• Area of parallelogram with adjacent sides $\vec{a}, \vec{b}$: $$|\vec{a} \times \vec{b}|$$
• Area of parallelogram with diagonals $\vec{d}_1, \vec{d}_2$: $$\frac{1}{2}|\vec{d}_1 \times \vec{d}_2|$$

5. Lagrange's Identity

$$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$ This identity connects the dot product, cross product, and magnitudes of two vectors.

Example 1: Magnitude of a Vector

Find the magnitude of the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$.

Solution: Step 1: Identify the scalar components of the vector. Here, $$x=1, \quad y=1, \quad z=1.$$

Step 2: Use the magnitude formula: $$|\vec{a}| = \sqrt{x^2+y^2+z^2}.$$

Step 3: Substitute the values: $$|\vec{a}| = \sqrt{1^2+1^2+1^2} = \sqrt{1+1+1} = \sqrt{3}.$$

Answer: $\sqrt{3}$

Example 2: Unit Vector

Find the unit vector in the direction of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$.

Solution: Step 1: Find the magnitude of the vector: $$|\vec{a}| = \sqrt{2^2+3^2+1^2} = \sqrt{4+9+1} = \sqrt{14}.$$

Step 2: A unit vector in the direction of $\vec{a}$ is obtained by dividing the vector by its magnitude: $$\hat{a} = \frac{\vec{a}}{|\vec{a}|}.$$

Step 3: Substitute the vector and simplify: $$\hat{a} = \frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}} = \frac{2}{\sqrt{14}}\hat{i}+\frac{3}{\sqrt{14}}\hat{j}+\frac{1}{\sqrt{14}}\hat{k}.$$

Answer: $\frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}$

Example 3: Direction Cosines

Find the direction cosines of the vector $\hat{i} + 2\hat{j} + 3\hat{k}$.

Solution: Step 1: The given vector is $$\vec{a} = \hat{i}+2\hat{j}+3\hat{k},$$ so its direction ratios are $$1, 2, 3.$$

Step 2: Find its magnitude: $$|\vec{a}| = \sqrt{1^2+2^2+3^2} = \sqrt{1+4+9} = \sqrt{14}.$$

Step 3: Direction cosines are given by dividing each direction ratio by the magnitude: $$l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}}.$$

These satisfy $$l^2+m^2+n^2 = \frac{1+4+9}{14}=1,$$ which verifies the result.

Answer: $1/\sqrt{14}, 2/\sqrt{14}, 3/\sqrt{14}$

Example 4: Sum of Vectors

Find the sum of the vectors $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$, and $\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}$.

Solution: Step 1: Add corresponding components: $$\vec{a}+\vec{b}+\vec{c} = (1-2+1)\hat{i} + (-2+4-6)\hat{j} + (1+5-7)\hat{k}.$$

Step 2: Simplify each component: $$= 0\hat{i} - 4\hat{j} - \hat{k}.$$

Step 3: Since the coefficient of $\hat{i}$ is zero, it may be omitted.

Answer: $-4\hat{j} - \hat{k}$

Example 5: Collinear Vectors

Show that the vectors $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} - 8\hat{k}$ are collinear.

Solution: Step 1: Let $$\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}, \qquad \vec{b} = -4\hat{i} + 6\hat{j} - 8\hat{k}.$$

Step 2: Check whether one vector is a scalar multiple of the other: $$\vec{b} = -2(2\hat{i} - 3\hat{j} + 4\hat{k}) = -2\vec{a}.$$

Step 3: Since one vector is a scalar multiple of the other, the vectors are parallel or collinear.

Answer: Proved.

Example 6: Internal Section Formula

Find the position vector of a point $R$ dividing the line segment joining $P(\hat{i} + 2\hat{j} - \hat{k})$ and $Q(-\hat{i} + \hat{j} + \hat{k})$ in the ratio 2:1 internally.

Solution: Step 1: Write the position vectors: $$\vec{p} = \hat{i}+2\hat{j}-\hat{k}, \qquad \vec{q} = -\hat{i}+\hat{j}+\hat{k}.$$ Given ratio: $$PR:RQ = 2:1.$$

Step 2: Use the internal section formula: $$\vec{r} = \frac{2\vec{q}+1\vec{p}}{2+1}.$$

Step 3: Substitute the vectors: $$\vec{r} = \frac{2(-\hat{i}+\hat{j}+\hat{k}) + (\hat{i}+2\hat{j}-\hat{k})}{3}.$$

Step 4: Simplify the numerator: $$= \frac{-2\hat{i}+2\hat{j}+2\hat{k}+\hat{i}+2\hat{j}-\hat{k}}{3} = \frac{-\hat{i}+4\hat{j}+\hat{k}}{3}.$$

Step 5: Separate the components: $$\vec{r} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}.$$

Answer: $-\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$

Example 7: External Section Formula

Find the position vector of a point $R$ dividing the line segment joining $P(\hat{i} + 2\hat{j} - \hat{k})$ and $Q(-\hat{i} + \hat{j} + \hat{k})$ in the ratio 2:1 externally.

Solution: Step 1: Use the external division formula: $$\vec{r} = \frac{2\vec{q} - 1\vec{p}}{2-1}.$$

Step 2: Substitute the vectors: $$\vec{r} = \frac{2(-\hat{i}+\hat{j}+\hat{k}) - (\hat{i}+2\hat{j}-\hat{k})}{1}.$$

Step 3: Simplify: $$\vec{r} = -2\hat{i}+2\hat{j}+2\hat{k}-\hat{i}-2\hat{j}+\hat{k} = -3\hat{i}+3\hat{k}.$$

Answer: $-3\hat{i} + 3\hat{k}$

Example 8: Midpoint Formula

Find the position vector of the midpoint of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

Solution: Step 1: Write the position vectors of the points: $$\vec{p} = 2\hat{i}+3\hat{j}+4\hat{k}, \qquad \vec{q} = 4\hat{i}+\hat{j}-2\hat{k}.$$

Step 2: Use the midpoint formula: $$\vec{r} = \frac{\vec{p}+\vec{q}}{2}.$$

Step 3: Add the vectors: $$\vec{r} = \frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2} = \frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}.$$

Step 4: Divide by 2: $$\vec{r} = 3\hat{i}+2\hat{j}+\hat{k}.$$

Answer: $3\hat{i} + 2\hat{j} + \hat{k}$

Example 9: Basic Dot Product

Find $\vec{a} \cdot \vec{b}$ if $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j} + 4\hat{k}$.

Solution: Step 1: Use the component formula for dot product: $$\vec{a} \cdot \vec{b} = (2)(-1) + (3)(2) + (-1)(4).$$

Step 2: Evaluate: $$\vec{a} \cdot \vec{b} = -2 + 6 - 4 = 0.$$

Step 3: Since the dot product is zero and the vectors are non-zero, the vectors are perpendicular.

Answer: 0

Example 10: Angle Between Two Vectors

Find the angle between the vectors $\hat{i} - 2\hat{j} + 3\hat{k}$ and $3\hat{i} - 2\hat{j} + \hat{k}$.

Solution: Step 1: Let $$\vec{a} = \hat{i}-2\hat{j}+3\hat{k}, \qquad \vec{b} = 3\hat{i}-2\hat{j}+\hat{k}.$$

Step 2: Find the dot product: $$\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3+4+3 = 10.$$

Step 3: Find the magnitudes: $$|\vec{a}| = \sqrt{1^2+(-2)^2+3^2} = \sqrt{14},$$ $$|\vec{b}| = \sqrt{3^2+(-2)^2+1^2} = \sqrt{14}.$$

Step 4: Use the formula for the angle between vectors: $$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{10}{\sqrt{14}\cdot\sqrt{14}} = \frac{10}{14} = \frac{5}{7}.$$

Therefore, $$\theta = \cos^{-1}\left(\frac{5}{7}\right).$$

Answer: $\cos^{-1}(5/7)$

Example 11: Projection of a Vector

Find the projection of the vector $\hat{i} + 3\hat{j} + 7\hat{k}$ on the vector $7\hat{i} - \hat{j} + 8\hat{k}$.

Solution: Step 1: Let $$\vec{a} = \hat{i}+3\hat{j}+7\hat{k}, \qquad \vec{b} = 7\hat{i}-\hat{j}+8\hat{k}.$$ The scalar projection of $\vec{a}$ on $\vec{b}$ is $$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}.$$

Step 2: Compute the dot product: $$\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7-3+56 = 60.$$

Step 3: Compute the magnitude of $\vec{b}$: $$|\vec{b}| = \sqrt{7^2+(-1)^2+8^2} = \sqrt{49+1+64} = \sqrt{114}.$$

Step 4: Therefore, $$\text{Projection} = \frac{60}{\sqrt{114}}.$$

Answer: $60/\sqrt{114}$

Example 12: Finding an Unknown in Orthogonal Vectors

Find $\lambda$ so that the vectors $2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\hat{i} - 2\hat{j} + 3\hat{k}$ are perpendicular to each other.

Solution: Step 1: Two vectors are perpendicular if their dot product is zero.

Step 2: Compute the dot product: $$(2\hat{i}+\lambda\hat{j}+\hat{k}) \cdot (\hat{i}-2\hat{j}+3\hat{k}) = 0.$$

Step 3: Multiply corresponding components: $$(2)(1) + (\lambda)(-2) + (1)(3) = 0.$$ $$2 - 2\lambda + 3 = 0.$$ $$5 - 2\lambda = 0.$$ $$2\lambda = 5 \implies \lambda = \frac{5}{2}.$$

Answer: $5/2$

Example 13: Using Dot Product Identities

If $|\vec{a}| = 2$, $|\vec{b}| = 3$, and $\vec{a} \cdot \vec{b} = 4$, find $|\vec{a} - \vec{b}|$.

Solution: Step 1: Use the identity $$|\vec{a}-\vec{b}|^2 = (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2.$$

Step 2: Substitute the given values: $$|\vec{a}-\vec{b}|^2 = 2^2 - 2(4) + 3^2 = 4 - 8 + 9 = 5.$$

Step 3: Take square root: $$|\vec{a}-\vec{b}| = \sqrt{5}.$$

Answer: $\sqrt{5}$

Example 14: Basic Cross Product

Find $\vec{a} \times \vec{b}$ if $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}$.

Solution: Step 1: Use determinant form: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix}.$$

Step 2: Expand along the first row: $$= \hat{i}[(1)(-2) - (3)(5)] - \hat{j}[(2)(-2) - (3)(3)] + \hat{k}[(2)(5) - (1)(3)].$$

Step 3: Simplify: $$= \hat{i}(-2-15) - \hat{j}(-4-9) + \hat{k}(10-3)$$ $$= -17\hat{i} + 13\hat{j} + 7\hat{k}.$$

Answer: $-17\hat{i} + 13\hat{j} + 7\hat{k}$

Example 15: Unit Normal Vector

Find a unit vector perpendicular to both $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Solution: Step 1: Compute the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix}.$$

Step 2: Expand: $$= \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}.$$

Step 3: Find the magnitude: $$|\vec{a} \times \vec{b}| = \sqrt{1^2+(-2)^2+1^2} = \sqrt{6}.$$

Step 4: Divide by the magnitude to get a unit vector: $$\hat{n} = \pm \frac{\hat{i}-2\hat{j}+\hat{k}}{\sqrt{6}}.$$

Answer: $\pm \frac{1}{\sqrt{6}}(\hat{i} - 2\hat{j} + \hat{k})$

Example 16: Area of a Parallelogram

Find the area of the parallelogram whose adjacent sides are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

Solution: Step 1: Area of the parallelogram is $$|\vec{a} \times \vec{b}|.$$

Step 2: Compute the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}.$$

Expanding, $$= \hat{i}(1+4) - \hat{j}(3-4) + \hat{k}(-3-1) = 5\hat{i}+\hat{j}-4\hat{k}.$$

Step 3: Find its magnitude: $$|\vec{a} \times \vec{b}| = \sqrt{5^2+1^2+(-4)^2} = \sqrt{25+1+16} = \sqrt{42}.$$

Answer: $\sqrt{42}$ square units

Example 17: Area of a Triangle

Find the area of a triangle with vertices $A(1, 1, 1)$, $B(1, 2, 3)$, and $C(2, 3, 1)$.

Solution: Step 1: Find two sides of the triangle: $$\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = \hat{j}+2\hat{k},$$ $$\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = \hat{i}+2\hat{j}.$$

Step 2: Compute the cross product: $$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}.$$

Expanding, $$= \hat{i}(0-4) - \hat{j}(0-2) + \hat{k}(0-1) = -4\hat{i}+2\hat{j}-\hat{k}.$$

Step 3: Find magnitude: $$|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2+2^2+(-1)^2} = \sqrt{21}.$$

Step 4: Area of triangle is half of this: $$\text{Area} = \frac{1}{2}\sqrt{21}.$$

Answer: $\sqrt{21}/2$ square units

Example 18: Work Done by a Force

Find the work done by the force $\vec{F} = 3\hat{i} + 2\hat{j} - 4\hat{k}$ displacing a particle from $A(1, -1, 2)$ to $B(2, -1, 3)$.

Solution: Step 1: Find the displacement vector: $$\vec{d} = \vec{AB} = (2-1)\hat{i} + (-1+1)\hat{j} + (3-2)\hat{k} = \hat{i}+\hat{k}.$$

Step 2: Work done is given by $$W = \vec{F} \cdot \vec{d}.$$

Step 3: Compute the dot product: $$W = (3\hat{i}+2\hat{j}-4\hat{k}) \cdot (\hat{i}+0\hat{j}+\hat{k}) = 3(1)+2(0)+(-4)(1) = -1.$$

Answer: -1 units

Example 20: Lagrange's Identity

If $|\vec{a}| = 2$, $|\vec{b}| = 5$, and $|\vec{a} \times \vec{b}| = 8$, find $\vec{a} \cdot \vec{b}$.

Solution: Step 1: Use Lagrange's identity: $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2.$$

Step 2: Substitute values: $$8^2 + (\vec{a} \cdot \vec{b})^2 = 2^2 \times 5^2.$$ $$64 + (\vec{a} \cdot \vec{b})^2 = 100.$$

Step 3: Solve: $$(\vec{a} \cdot \vec{b})^2 = 36 \implies \vec{a} \cdot \vec{b} = \pm 6.$$

The sign cannot be fixed with the given information alone.

Answer: $\pm 6$

JEE Main Standard

Example 21: Vector Sum to Zero

If $\vec{a}, \vec{b}, \vec{c}$ are vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{c}| = 5$, find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Solution: Step 1: Square both sides of $$\vec{a}+\vec{b}+\vec{c}=\vec{0}.$$ This means $$(\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = 0.$$

Step 2: Expand: $$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0.$$

Step 3: Substitute the magnitudes: $$3^2 + 4^2 + 5^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0.$$ $$9+16+25 + 2S = 0,$$ where $$S = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}.$$

Step 4: Solve: $$50 + 2S = 0 \implies 2S = -50 \implies S = -25.$$

Answer: -25

Example 22: Angle between Vectors

If $\vec{a}$ and $\vec{b}$ are unit vectors and $\theta$ is the angle between them, prove that $\sin(\theta/2) = \frac{1}{2}|\vec{a} - \vec{b}|$.

Solution: Step 1: Start with $$|\vec{a}-\vec{b}|^2 = (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}).$$

Step 2: Expand: $$|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}).$$

Step 3: Since $\vec{a}$ and $\vec{b}$ are unit vectors, $$|\vec{a}|=|\vec{b}|=1,$$ and $$\vec{a} \cdot \vec{b} = \cos\theta.$$ So, $$|\vec{a}-\vec{b}|^2 = 1+1-2\cos\theta = 2(1-\cos\theta).$$

Step 4: Use the identity $$1-\cos\theta = 2\sin^2(\theta/2).$$ Thus, $$|\vec{a}-\vec{b}|^2 = 2\cdot 2\sin^2(\theta/2) = 4\sin^2(\theta/2).$$

Step 5: Taking positive square roots, $$|\vec{a}-\vec{b}| = 2\sin(\theta/2).$$ Hence, $$\sin(\theta/2) = \frac{1}{2}|\vec{a}-\vec{b}|.$$

Answer: Proved.

Example 23: Cross Product Condition

If $\vec{a} \times \vec{b} = \vec{c} \times \vec{d}$ and $\vec{a} \times \vec{c} = \vec{b} \times \vec{d}$, show that $(\vec{a} - \vec{d})$ is parallel to $(\vec{b} - \vec{c})$.

Solution: Step 1: To prove two vectors are parallel, it is enough to show that their cross product is zero. So consider $$(\vec{a}-\vec{d}) \times (\vec{b}-\vec{c}).$$

Step 2: Expand: $$= \vec{a} \times \vec{b} - \vec{a} \times \vec{c} - \vec{d} \times \vec{b} + \vec{d} \times \vec{c}.$$

Step 3: Use anti-commutativity: $$-\vec{d} \times \vec{b} = \vec{b} \times \vec{d}, \qquad \vec{d} \times \vec{c} = -\vec{c} \times \vec{d}.$$ So, $$= \vec{a} \times \vec{b} - \vec{a} \times \vec{c} + \vec{b} \times \vec{d} - \vec{c} \times \vec{d}.$$

Step 4: Use the given conditions: $$\vec{a} \times \vec{b} = \vec{c} \times \vec{d}, \qquad \vec{b} \times \vec{d} = \vec{a} \times \vec{c}.$$ Thus, $$= (\vec{c} \times \vec{d}) - \vec{a} \times \vec{c} + \vec{a} \times \vec{c} - (\vec{c} \times \vec{d}) = \vec{0}.$$

Step 5: Since the cross product is zero, $$(\vec{a}-\vec{d}) \parallel (\vec{b}-\vec{c}).$$

Answer: Proved.

Example 24: Collinearity using Cross Product

Find the value of $x$ for which the points $A(3, 2, 1)$, $B(4, x, 5)$, and $C(4, 2, -2)$ are collinear.

Solution: Step 1: For collinearity, vectors $\vec{AB}$ and $\vec{AC}$ must be parallel.

Step 2: Compute the vectors: $$\vec{AB} = (4-3)\hat{i} + (x-2)\hat{j} + (5-1)\hat{k} = \hat{i} + (x-2)\hat{j} + 4\hat{k},$$ $$\vec{AC} = (4-3)\hat{i} + (2-2)\hat{j} + (-2-1)\hat{k} = \hat{i} - 3\hat{k}.$$

Step 3: For collinearity, one vector must be a scalar multiple of the other. From the $\hat{i}$-components, $$1 = \lambda(1) \implies \lambda = 1.$$ Then from the $\hat{j}$-components, $$x-2 = \lambda(0) = 0 \implies x=2.$$ But from the $\hat{k}$-components, $$4 = \lambda(-3) = -3,$$ which is impossible.

Step 4: Hence no value of $x$ can make these three points collinear.

Answer: No such value of $x$ exists.

Example 25: Mutual Orthogonality

If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitude, find the angle $\vec{a} + \vec{b} + \vec{c}$ makes with $\vec{a}$.

Solution: Step 1: Let $$|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda,$$ and since they are mutually perpendicular, $$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0.$$

Step 2: Let $$\vec{r} = \vec{a}+\vec{b}+\vec{c}.$$ Find its magnitude: $$|\vec{r}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 = 3\lambda^2,$$ so $$|\vec{r}| = \lambda\sqrt{3}.$$

Step 3: Let the angle between $\vec{r}$ and $\vec{a}$ be $\theta$. Then $$\cos\theta = \frac{\vec{r} \cdot \vec{a}}{|\vec{r}| |\vec{a}|}.$$

Step 4: Compute the numerator: $$\vec{r} \cdot \vec{a} = (\vec{a}+\vec{b}+\vec{c})\cdot\vec{a} = |\vec{a}|^2+0+0 = \lambda^2.$$

Step 5: Therefore, $$\cos\theta = \frac{\lambda^2}{(\lambda\sqrt{3})(\lambda)} = \frac{1}{\sqrt{3}}.$$ Thus, $$\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right).$$

Answer: $\cos^{-1}(1/\sqrt{3})$

Example 26: Linear Combination Solution

Find a vector $\vec{d}$ which is perpendicular to both $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, and satisfies $\vec{c} \cdot \vec{d} = 15$, where $\vec{c} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Solution: Step 1: A vector perpendicular to both $\vec{a}$ and $\vec{b}$ must be parallel to $$\vec{a} \times \vec{b}.$$ Compute it: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix}.$$

Expanding, $$= \hat{i}((-1)(1)-1(1)) - \hat{j}(1\cdot1-1\cdot1) + \hat{k}(1\cdot1-(-1)(1))$$ $$= -2\hat{i} + 0\hat{j} + 2\hat{k}.$$

Step 2: So let $$\vec{d} = \lambda(-2\hat{i}+2\hat{k}) = -2\lambda\hat{i}+2\lambda\hat{k}.$$

Step 3: Use the condition $$\vec{c} \cdot \vec{d} = 15.$$ That is, $$(\hat{i}+2\hat{j}+3\hat{k}) \cdot (-2\lambda\hat{i}+0\hat{j}+2\lambda\hat{k}) = 15.$$

Step 4: Compute the dot product: $$-2\lambda + 0 + 6\lambda = 15 \implies 4\lambda = 15 \implies \lambda = \frac{15}{4}.$$

Step 5: Therefore, $$\vec{d} = -2\left(\frac{15}{4}\right)\hat{i} + 2\left(\frac{15}{4}\right)\hat{k} = -\frac{15}{2}\hat{i} + \frac{15}{2}\hat{k}.$$

Answer: $-\frac{15}{2}\hat{i} + \frac{15}{2}\hat{k}$

Example 27: Torque (Moment of a Force)

Find the torque (moment of a force) $\vec{\tau}$ of a force $\vec{F} = 3\hat{i} + 2\hat{j} - 4\hat{k}$ acting at the point $P(1, -1, 2)$ about the point $A(2, -1, 3)$.

Solution: Step 1: Torque about point $A$ is given by $$\vec{\tau} = \vec{r} \times \vec{F},$$ where $\vec{r}$ is the position vector of the point of application relative to the point about which moment is taken.

Step 2: Find $$\vec{r} = \vec{AP} = (1-2)\hat{i} + (-1+1)\hat{j} + (2-3)\hat{k} = -\hat{i} - \hat{k}.$$

Step 3: Compute the cross product: $$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 0 & -1 \\ 3 & 2 & -4 \end{vmatrix}.$$

Expanding, $$= \hat{i}(0-(-2)) - \hat{j}(4-(-3)) + \hat{k}(-2-0)$$ $$= 2\hat{i} - 7\hat{j} - 2\hat{k}.$$

Answer: $2\hat{i} - 7\hat{j} - 2\hat{k}$

Example 28: Parallelogram Area via Diagonals

The diagonals of a parallelogram are represented by the vectors $\vec{d}_1 = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d}_2 = \hat{i} - 3\hat{j} + 4\hat{k}$. Find its area.

Solution: Step 1: Area of a parallelogram in terms of diagonals is $$\text{Area} = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2|.$$

Step 2: Compute the cross product: $$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}.$$

Expanding, $$= \hat{i}(4-6) - \hat{j}(12-(-2)) + \hat{k}(-9-1)$$ $$= -2\hat{i} - 14\hat{j} - 10\hat{k}.$$

Step 3: Find its magnitude: $$|\vec{d}_1 \times \vec{d}_2| = \sqrt{(-2)^2+(-14)^2+(-10)^2} = \sqrt{300} = 10\sqrt{3}.$$

Step 4: Therefore, $$\text{Area} = \frac{1}{2}(10\sqrt{3}) = 5\sqrt{3}.$$

Answer: $5\sqrt{3}$ square units.

Example 29: Equality of Dot and Cross Products

If $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$ and $\vec{a} \times \vec{b} = \vec{a} \times \vec{c}$, and $\vec{a}$ is not the zero vector, show that $\vec{b} = \vec{c}$.

Solution: Step 1: From $$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c},$$ we get $$\vec{a} \cdot (\vec{b}-\vec{c}) = 0.$$ So $\vec{a}$ is perpendicular to $(\vec{b}-\vec{c})$.

Step 2: From $$\vec{a} \times \vec{b} = \vec{a} \times \vec{c},$$ we get $$\vec{a} \times (\vec{b}-\vec{c}) = \vec{0}.$$ So $\vec{a}$ is parallel to $(\vec{b}-\vec{c})$.

Step 3: A non-zero vector cannot be both parallel and perpendicular to another non-zero vector. Therefore the only possibility is $$\vec{b}-\vec{c} = \vec{0}.$$ Hence, $$\vec{b} = \vec{c}.$$

Answer: Proved.

Example 30: Angle between Diagonals

Let $\vec{a}$ and $\vec{b}$ be adjacent sides of a parallelogram. Find the angle between the diagonals if $\vec{a} = 2\hat{i} + \hat{j}$ and $\vec{b} = \hat{i} + 2\hat{j}$.

Solution: Step 1: The diagonals of the parallelogram are $$\vec{d}_1 = \vec{a}+\vec{b}, \qquad \vec{d}_2 = \vec{a}-\vec{b}.$$

Step 2: Compute the diagonals: $$\vec{d}_1 = (2+1)\hat{i} + (1+2)\hat{j} = 3\hat{i}+3\hat{j},$$ $$\vec{d}_2 = (2-1)\hat{i} + (1-2)\hat{j} = \hat{i}-\hat{j}.$$

Step 3: Find their dot product: $$\vec{d}_1 \cdot \vec{d}_2 = (3)(1) + (3)(-1) = 3-3 = 0.$$

Step 4: Since the dot product is zero, the diagonals are perpendicular. Hence the angle between them is $$90^\circ.$$

Answer: $90^\circ$