Q.Find the angle between force $\vec{F} = (3\hat{i} + 4\hat{j} - 5\hat{k})$ unit and displacement $\vec{d} = (5\hat{i} + 4\hat{j} + 3\hat{k})$ unit. Also find the projection of $\vec{F}$ on $\vec{d}$.

Concept: Dot Product for Angle and Projection

1. First, calculate the dot product of the vectors and their magnitudes:

   $\vec{F} \cdot \vec{d} = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16$

   $|\vec{F}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$

   $|\vec{d}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$

2. The angle $\theta$ between $\vec{F}$ and $\vec{d}$ is found using $\cos\theta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}| |\vec{d}|}$:

   $\cos\theta = \frac{16}{(5\sqrt{2})(5\sqrt{2})} = \frac{16}{50} = \frac{8}{25}$

   $\theta = \cos^{-1}\left(\frac{8}{25}\right)$ …

To find the angle between two vectors and the projection of one vector onto another, the dot product is our fundamental tool. The dot product, also known as the scalar product, provides a way to relate the algebraic components of vectors to their geometric relationship, specifically the angle between them.

Finding the Angle Between Vectors

The dot product of two vectors $\vec{A}$ and $\vec{B}$ can be defined in two ways:

1. Algebraically: If $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$, then $\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$.
2. Geometrically: $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta$, where $|\vec{A}|$ and $|\vec{B}|$ are the magnitudes of the vectors, and $\theta$ is the angle between them.

By equating these two definitions, we get a powerful formula to find the angle:

Let's apply this to our problem:

1. Identify the vectors.

   We are given the force vector $\vec{F}$ and the displacement vector $\vec{d}$:

   $\vec{F} = 3\hat{i} + 4\hat{j} - 5\hat{k}$

   $\vec{d} = 5\hat{i} + 4\hat{j} + 3\hat{k}$

2. Calculate the dot product $\vec{F} \cdot \vec{d}$.

   We multiply the corresponding components and sum them up:

   $\vec{F} \cdot \vec{d} = (3)(5) + (4)(4) + (-5)(3)$

   $\vec{F} \cdot \vec{d} = 15 + 16 - 15$

   $\vec{F} \cdot \vec{d} = 16$

3. Calculate the magnitudes of $\vec{F}$ and $\vec{d}$.

   The magnitude of a vector $\vec{V} = V_x\hat{i} + V_y\hat{j} + V_z\hat{k}$ is given by $|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$.

   For $\vec{F}$:

   $|\vec{F}| = \sqrt{3^2 + 4^2 + (-5)^2}$

   $|\vec{F}| = \sqrt{9 + 16 + 25}$

   $|\vec{F}| = \sqrt{50}$

   $|\vec{F}| = 5\sqrt{2}$

   For $\vec{d}$:

   $|\vec{d}| = \sqrt{5^2 + 4^2 + 3^2}$

   $|\vec{d}| = \sqrt{25 + 16 + 9}$

   $|\vec{d}| = \sqrt{50}$

   $|\vec{d}| = 5\sqrt{2}$

   Tip

   Notice that both vectors have the same magnitude, $\sqrt{50}$. This simplifies calculations slightly.

4. Apply the dot product formula to find $\cos\theta$.

   Substitute the calculated dot product and magnitudes into the formula $\cos\theta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}||\vec{d}|}$:

   $\cos\theta = \frac{16}{(5\sqrt{2})(5\sqrt{2})}$

   $\cos\theta = \frac{16}{25 \cdot 2}$

   $\cos\theta = \frac{16}{50}$

   $\cos\theta = \frac{8}{25}$

5. Find the angle $\theta$.

   To find $\theta$, we take the inverse cosine (arccosine) of the value:

   $\theta = \arccos\left(\frac{8}{25}\right)$

Finding the Projection of $\vec{F}$ on $\vec{d}$ …

Concept: Vector Dot Product — Angle and Projection

Step 1: Write the vectors and compute the dot product.

Step 2: Compute the magnitudes.

Step 3: Find the angle using $\cos\theta = \dfrac{\vec{F}\cdot\vec{d}}{|\vec{F}||\vec{d}|}$. …