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Given the following vectors d = 4i + 5j + 7k and b = 2i + 5j + 4k, find (i) d x b (ii) b.b (iii) the angle between them.

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3 parts
a

−23i⃗−2j⃗+14k⃗-23\vec{i} - 2\vec{j} + 14\vec{k}

b
51
c

27.88∘27.88^\circ

Step-by-step solution

11 steps

let's tackle these vector calculations.

Given the vectors:
d⃗=4i⃗+3j⃗+7k⃗\vec{d} = 4\vec{i} + 3\vec{j} + 7\vec{k}
e⃗=2i⃗+5j⃗+4k⃗\vec{e} = 2\vec{i} + 5\vec{j} + 4\vec{k}

Part a· 4 steps

Determine d⃗×e⃗\vec{d} \times \vec{e}

  1. Set up the determinant for the cross product.

    d⃗×e⃗=∣i⃗j⃗k⃗437254∣\vec{d} \times \vec{e} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & 3 & 7 \\ 2 & 5 & 4 \end{vmatrix}

  2. Expand the determinant.

    d⃗×e⃗=i⃗((3)(4)−(7)(5))−j⃗((4)(4)−(7)(2))+k⃗((4)(5)−(3)(2))\vec{d} \times \vec{e} = \vec{i}((3)(4) - (7)(5)) - \vec{j}((4)(4) - (7)(2)) + \vec{k}((4)(5) - (3)(2))

  3. Perform the multiplications and subtractions.

    d⃗×e⃗=i⃗(12−35)−j⃗(16−14)+k⃗(20−6)\vec{d} \times \vec{e} = \vec{i}(12 - 35) - \vec{j}(16 - 14) + \vec{k}(20 - 6)

  4. Simplify to find the resulting vector.

    d⃗×e⃗=−23i⃗−2j⃗+14k⃗\vec{d} \times \vec{e} = \boxed{-23\vec{i} - 2\vec{j} + 14\vec{k}}

Part b51

Determine d⃗⋅e⃗\vec{d} \cdot \vec{e}

  1. Use the dot product formula d⃗⋅e⃗=dxex+dyey+dzez\vec{d} \cdot \vec{e} = d_x e_x + d_y e_y + d_z e_z.

    d⃗⋅e⃗=(4)(2)+(3)(5)+(7)(4)\vec{d} \cdot \vec{e} = (4)(2) + (3)(5) + (7)(4)

  2. Perform the multiplications.

    d⃗⋅e⃗=8+15+28\vec{d} \cdot \vec{e} = 8 + 15 + 28

  3. Sum the values.

    d⃗⋅e⃗=51\vec{d} \cdot \vec{e} = \boxed{51}

Part c

27.88∘27.88^\circ

Determine the angle between them

  1. Calculate the magnitudes of vectors d⃗\vec{d} and e⃗\vec{e}.

    ∣d⃗∣=42+32+72=16+9+49=74|\vec{d}| = \sqrt{4^2 + 3^2 + 7^2} = \sqrt{16 + 9 + 49} = \sqrt{74} ∣e⃗∣=22+52+42=4+25+16=45|\vec{e}| = \sqrt{2^2 + 5^2 + 4^2} = \sqrt{4 + 25 + 16} = \sqrt{45}

  2. Use the dot product formula d⃗⋅e⃗=∣d⃗∣∣e⃗∣cos⁡θ\vec{d} \cdot \vec{e} = |\vec{d}| |\vec{e}| \cos\theta to find cos⁡θ\cos\theta.

    cos⁡θ=d⃗⋅e⃗∣d⃗∣∣e⃗∣=517445\cos\theta = \frac{\vec{d} \cdot \vec{e}}{|\vec{d}| |\vec{e}|} = \frac{51}{\sqrt{74}\sqrt{45}}

  3. Calculate the product of the square roots and simplify.

    cos⁡θ=5174×45=513330\cos\theta = \frac{51}{\sqrt{74 \times 45}} = \frac{51}{\sqrt{3330}} cos⁡θ≈5157.706≈0.8838\cos\theta \approx \frac{51}{57.706} \approx 0.8838

  4. Find the angle θ\theta by taking the inverse cosine.

    θ=arccos⁡(0.8838)\theta = \arccos(0.8838) θ≈27.88∘\theta \approx \boxed{27.88^\circ}
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