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Worked Examples · Example 5.1

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

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We use the dot product to find the angle between the vectors, yielding θ=arccos⁡(825)\theta = \arccos\left(\frac{8}{25}\right), and then apply the projection formula to find the projection of F⃗\vec{F} on d⃗\vec{d} as 825\frac{8\sqrt{2}}{5} units.

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 A⃗\vec{A} and B⃗\vec{B} can be defined in two ways:

  1. Algebraically: If A⃗=Axi^+Ayj^+Azk^\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} and B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}, then A⃗⋅B⃗=AxBx+AyBy+AzBz\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z.
  2. Geometrically: A⃗⋅B⃗=∣A⃗∣∣B⃗∣cos⁡θ\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta, where ∣A⃗∣|\vec{A}| and ∣B⃗∣|\vec{B}| are the magnitudes of the vectors, and θ\theta is the angle between them.

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

cos⁡θ=A⃗⋅B⃗∣A⃗∣∣B⃗∣\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}

This formula allows us to calculate the cosine of the angle using only the components of the vectors.

Let's apply this to our problem:

  1. Identify the vectors.

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

    F⃗=3i^+4j^−5k^\vec{F} = 3\hat{i} + 4\hat{j} - 5\hat{k}

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

  2. Calculate the dot product F⃗⋅d⃗\vec{F} \cdot \vec{d}.

    We multiply the corresponding components and sum them up:

    F⃗⋅d⃗=(3)(5)+(4)(4)+(−5)(3)\vec{F} \cdot \vec{d} = (3)(5) + (4)(4) + (-5)(3)

    F⃗⋅d⃗=15+16−15\vec{F} \cdot \vec{d} = 15 + 16 - 15

    F⃗⋅d⃗=16\vec{F} \cdot \vec{d} = 16

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

    The magnitude of a vector V⃗=Vxi^+Vyj^+Vzk^\vec{V} = V_x\hat{i} + V_y\hat{j} + V_z\hat{k} is given by ∣V⃗∣=Vx2+Vy2+Vz2|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}.

    For F⃗\vec{F}:

    ∣F⃗∣=32+42+(−5)2|\vec{F}| = \sqrt{3^2 + 4^2 + (-5)^2}

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

    ∣F⃗∣=50|\vec{F}| = \sqrt{50}

    ∣F⃗∣=52|\vec{F}| = 5\sqrt{2}

    For d⃗\vec{d}:

    ∣d⃗∣=52+42+32|\vec{d}| = \sqrt{5^2 + 4^2 + 3^2}

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

    ∣d⃗∣=50|\vec{d}| = \sqrt{50}

    ∣d⃗∣=52|\vec{d}| = 5\sqrt{2}

    Tip

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

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

    Substitute the calculated dot product and magnitudes into the formula cos⁡θ=F⃗⋅d⃗∣F⃗∣∣d⃗∣\cos\theta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}||\vec{d}|}:

    cos⁡θ=16(52)(52)\cos\theta = \frac{16}{(5\sqrt{2})(5\sqrt{2})}

    cos⁡θ=1625⋅2\cos\theta = \frac{16}{25 \cdot 2}

    cos⁡θ=1650\cos\theta = \frac{16}{50}

    cos⁡θ=825\cos\theta = \frac{8}{25}

  5. Find the angle θ\theta.

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

    θ=arccos⁡(825)\theta = \arccos\left(\frac{8}{25}\right)

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

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