For the points P(−2, 1) and Q(3, 5), in terms of i and j find: a) PQ, b) |PQ|, c) PQ

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Here's the solution for question 5. 5. For the points P(−2, 1) and Q(3, 5), in terms of i and j find: First, express the position vectors of P and Q: P = −2i + 1j; Q = 3i + 5j a) PQ Step 1: To find the vector PQ, subtract the position vector of P from the position vector of Q. PQ = Q − P = (3i + 5j) − (−2i + 1j) Step 2: Distribute the negative sign and combine the i and j components. PQ = (3 − (−2))i + (5 − 1)j; PQ = (3 + 2)i + (4)j; PQ = 5i + 4j The vector PQ is 5i + 4j. b) |PQ| Step 1: To find the magnitude of PQ, use the formula |PQ| = √(x² + y²), where PQ = xi + yj. From part (a), x=5 and y=4. |PQ| = √((5)² + (4)²) Step 2: Calculate the squares and sum them. |PQ| = √(25 + 16); |PQ| = √(41) The magnitude |PQ| is √(41). c) PQ Step 1: To find the unit vector PQ in the direction of PQ, divide the vector PQ by its magnitude |PQ|. PQ = PQ|PQ| = 5i + 4j√(41) Step 2: Express the unit vector with each component divided by the magnitude. PQ = (5)/(√(41))i + (4)/(√(41))j The unit vector PQ is (5)/(√(41))i + (4)/(√(41))j.

For the points P(-2, 1) and Q(3, 5), in terms of i and j find: a) PQ, b) |PQ|, c) PQ

Step-by-step solution

Part a· 2 steps

To find the vector $\vec{PQ}$ , subtract the position vector of P from the position vector of Q.

Distribute the negative sign and combine the $\mathbf{i}$ and $\mathbf{j}$ components.

Part b· 2 steps

To find the magnitude of $\vec{PQ}$ , use the formula $|\vec{PQ}| = \sqrt{x^2 + y^2}$ , where $\vec{PQ} = x\mathbf{i} + y\mathbf{j}$ .

Calculate the squares and sum them.

Part c· 2 steps

To find the unit vector $\widehat{PQ}$ in the direction of $\vec{PQ}$ , divide the vector $\vec{PQ}$ by its magnitude $|\vec{PQ}|$ .

Express the unit vector with each component divided by the magnitude.

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