Quick Solution

Given:

$\vec{a} = \hat{i} - \hat{k}, \quad \vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$

Form the matrix:

$M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix}$

Find the determinant:

$\det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1$

Since the determinant is constant and non-zero, the vectors are linearly independent.

$\boxed{\text{The matrix does not depend on } x \text{ or } y}$

Quick Solution

Given: $\vec{a}, \vec{b}$ are unit vectors and

$2\vec{a} + \vec{b} = 3$

Take magnitude on both sides:

$|2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9$

Use identity:

$$|2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b})$$

Set equal to 9:

$$5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ$$

Vector Projection Problem

Given: A vector of magnitude 5 makes equal angles with x, y, and z axes.
To Find: Sum of magnitudes of projections on each axis.

Let angle with each axis be $\alpha$. Then, from direction cosine identity: $$\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \Rightarrow 3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \frac{1}{\sqrt{3}}$$

Projection on each axis: $5 \cdot \frac{1}{\sqrt{3}}$
Sum = $3 \cdot \frac{5}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \boxed{5\sqrt{3}}$

✅ Final Answer: $\boxed{5\sqrt{3}}$

Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

Vectors:

• $\vec{a} = m\hat{i} + \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} - \hat{j} + \hat{k}$
• $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$

Step 1: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$

First compute $\vec{b} \times \vec{c}$:

$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\ = \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k}$

Step 2: Compute dot product with $\vec{a}$:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5$

Step 3: Volume = $| -m + 5 | = 4$

So, $|-m + 5| = 4 \Rightarrow -m + 5 = \pm 4$

• Case 1: $-m + 5 = 4 \Rightarrow m = 1$
• Case 2: $-m + 5 = -4 \Rightarrow m = 9$

✅ Final Answer: $\boxed{m = 1 \text{ or } 9}$

Given: Vectors:

• $\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}$
• $\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}$

Condition: Vectors are coplanar ⟹ Scalar triple product = 0

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Step 1: Use determinant:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \lambda^2 & 1 & 1 \\ 1 & \lambda^2 & 1 \\ 1 & 1 & \lambda^2 \end{vmatrix}$

Step 2: Expand the determinant:

$= \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\ = \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2)$

Simplify:

$= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2$

Step 3: Set scalar triple product to 0:

$\lambda^6 - 3\lambda^2 + 2 = 0$

Step 4: Let $x = \lambda^2$, then:

$x^3 - 3x + 2 = 0$

Factor:

$x^3 - 3x + 2 = (x - 1)^2(x + 2)$

So, $\lambda^2 = 1$ (double root), or $\lambda^2 = -2$ (discard as it's not real)

Thus, real values of $\lambda$ are: $\lambda = \pm1$

✅ Final Answer: $\boxed{2}$ distinct real values

Formula for work done:

$$W = |F| \cdot |D| \cdot \cos\theta$$

Given:

• $|F| = 40 \, \text{N}$
• $|D| = 3 \, \text{m}$
• $\theta = 60^\circ$

Step 1: Plug in the values:

$$W = 40 \cdot 3 \cdot \cos(60^\circ)$$

Step 2: Use $\cos(60^\circ) = \frac{1}{2}$

$$W = 40 \cdot 3 \cdot \frac{1}{2} = 60 \, \text{J}$$

✅ Final Answer: $\boxed{60 \, \text{J}}$

A man starts at the origin $O$, walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point $P$. Find the displacement vector $\vec{OP}$.

? Solution:

• North-East (45°): $$\vec{A} = 3 \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right)$$
• North-West (135°): $$\vec{B} = 4 \cdot \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( -\frac{4}{\sqrt{2}}, \frac{4}{\sqrt{2}} \right)$$
• Total Displacement: $$\vec{OP} = \vec{A} + \vec{B} = \left( \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$$

✅ Final Answer:

$$\boxed{ \vec{OP} = \left( \frac{-1}{\sqrt{2}},\ \frac{7}{\sqrt{2}} \right) }$$

Work Done Problem:

A crate is pulled 25 m along a dock with a force of 180 N at an angle of 45°.

✅ Formula Used:

$$\text{Work} = F \cdot d \cdot \cos(\theta)$$

✅ Substituting Values:

$$W = 180 \times 25 \times \cos(45^\circ) = 180 \times 25 \times 0.70710678118 = 3181.98052\, \text{J}$$

✅ Final Answer (to 5 decimal places):

$$\boxed{3.181\times 10^3 \, \text{Joules}}$$