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Given vectors P = -2i + 3j, Q = 3i + 4j, S = 3i, calculate the angle 2220PSQ.

Mathematics

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Answer

59∘59^\circ

Step-by-step solution

6 steps

To calculate the angle ∠PSQ\angle PSQ, we need to find the vectors SP⃗\vec{SP} and SQ⃗\vec{SQ} and then use the dot product formula.

Given position vectors:
P⃗=−2i+3j\vec{P} = -2\mathbf{i} + 3\mathbf{j}
Q⃗=3i+4j\vec{Q} = 3\mathbf{i} + 4\mathbf{j}
S⃗=3i\vec{S} = 3\mathbf{i}

  1. Find the vector SP⃗\vec{SP}.

    SP⃗=P⃗−S⃗\vec{SP} = \vec{P} - \vec{S} SP⃗=(−2i+3j)−(3i)\vec{SP} = (-2\mathbf{i} + 3\mathbf{j}) - (3\mathbf{i}) SP⃗=(−2−3)i+(3−0)j\vec{SP} = (-2 - 3)\mathbf{i} + (3 - 0)\mathbf{j} SP⃗=−5i+3j\vec{SP} = -5\mathbf{i} + 3\mathbf{j}

  2. Find the vector SQ⃗\vec{SQ}.

    SQ⃗=Q⃗−S⃗\vec{SQ} = \vec{Q} - \vec{S} SQ⃗=(3i+4j)−(3i)\vec{SQ} = (3\mathbf{i} + 4\mathbf{j}) - (3\mathbf{i}) SQ⃗=(3−3)i+(4−0)j\vec{SQ} = (3 - 3)\mathbf{i} + (4 - 0)\mathbf{j} SQ⃗=0i+4j=4j\vec{SQ} = 0\mathbf{i} + 4\mathbf{j} = 4\mathbf{j}

  3. Calculate the dot product SP⃗⋅SQ⃗\vec{SP} \cdot \vec{SQ}.

    SP⃗⋅SQ⃗=(−5)(0)+(3)(4)\vec{SP} \cdot \vec{SQ} = (-5)(0) + (3)(4) SP⃗⋅SQ⃗=0+12=12\vec{SP} \cdot \vec{SQ} = 0 + 12 = 12

  4. Calculate the magnitudes of SP⃗\vec{SP} and SQ⃗\vec{SQ}.

    ∣SP⃗∣=(−5)2+(3)2=25+9=34|\vec{SP}| = \sqrt{(-5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34} ∣SQ⃗∣=(0)2+(4)2=0+16=16=4|\vec{SQ}| = \sqrt{(0)^2 + (4)^2} = \sqrt{0 + 16} = \sqrt{16} = 4

  5. Use the dot product formula cos⁡θ=a⋅b∣a∣∣b∣\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} to find ∠PSQ\angle PSQ.

    cos⁡(∠PSQ)=SP⃗⋅SQ⃗∣SP⃗∣∣SQ⃗∣\cos(\angle PSQ) = \frac{\vec{SP} \cdot \vec{SQ}}{|\vec{SP}| |\vec{SQ}|} cos⁡(∠PSQ)=1234×4\cos(\angle PSQ) = \frac{12}{\sqrt{34} \times 4} cos⁡(∠PSQ)=12434\cos(\angle PSQ) = \frac{12}{4\sqrt{34}} cos⁡(∠PSQ)=334\cos(\angle PSQ) = \frac{3}{\sqrt{34}}

  6. Calculate the angle to the nearest degree.

    ∠PSQ=arccos⁡(334)\angle PSQ = \arccos\left(\frac{3}{\sqrt{34}}\right) ∠PSQ≈arccos⁡(0.51450)\angle PSQ \approx \arccos(0.51450) ∠PSQ≈59.04∘\angle PSQ \approx 59.04^\circ

    Rounding to the nearest degree, the angle is 59∘59^\circ.

    The angle ∠PSQ\angle PSQ is 59∘\boxed{59^\circ}.

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