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Given that OP = (4, 1), where O is the origin and Q is the point (1;5), find |OP|. Leave the answer in surd form. Given also that QR = (-2, -3), find the coordinates of R and PR.

Mathematics

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Answers

3 parts
a

17\sqrt{17}

b(i)
(-1, 2)
b(ii)

(−51)\begin{pmatrix} -5 \\ 1 \end{pmatrix}

Step-by-step solution

5 steps

Here's the step-by-step solution for question 17.

Part a· 1 step

Find ∣OP⃗∣|\vec{OP}|. Leave the answer in surd form.

Given OP⃗=(41)\vec{OP} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}.

  1. Calculate the magnitude of the vector.

    The magnitude of a vector (xy)\begin{pmatrix} x \\ y \end{pmatrix} is given by x2+y2\sqrt{x^2 + y^2}.

    ∣OP⃗∣=42+12|\vec{OP}| = \sqrt{4^2 + 1^2} ∣OP⃗∣=16+1|\vec{OP}| = \sqrt{16 + 1} ∣OP⃗∣=17|\vec{OP}| = \sqrt{17}

    The magnitude of OP⃗\vec{OP} is 17\boxed{\sqrt{17}}.

Part b(i)(-1, 2)

the coordinates of R.

Given also that QR⃗=(−2−3)\vec{QR} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}, find

Given Q is the point (1;5)(1; 5). Let the coordinates of R be (xR,yR)(x_R, y_R).
The vector QR⃗\vec{QR} is given by (xR−xQyR−yQ)\begin{pmatrix} x_R - x_Q \\ y_R - y_Q \end{pmatrix}.

  1. Set up equations for the components.

    (xR−1yR−5)=(−2−3)\begin{pmatrix} x_R - 1 \\ y_R - 5 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \end{pmatrix}

    This gives two equations:
    xR−1=−2x_R - 1 = -2
    yR−5=−3y_R - 5 = -3

  2. Solve for xRx_R and yRy_R.

    xR=−2+1=−1x_R = -2 + 1 = -1
    yR=−3+5=2y_R = -3 + 5 = 2
    The coordinates of R are (−1,2)\boxed{(-1, 2)}.

Part b(ii)

(−51)\begin{pmatrix} -5 \\ 1 \end{pmatrix}

PR⃗\vec{PR}.

To find PR⃗\vec{PR}, we use the formula PR⃗=OR⃗−OP⃗\vec{PR} = \vec{OR} - \vec{OP}.
From part (b)(i), the coordinates of R are (−1,2)(-1, 2), so OR⃗=(−12)\vec{OR} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} (since O is the origin).
From the question, OP⃗=(41)\vec{OP} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}.

  1. Substitute the vectors into the formula.

    PR⃗=(−12)−(41)\vec{PR} = \begin{pmatrix} -1 \\ 2 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix}

  2. Perform the vector subtraction.

    PR⃗=(−1−42−1)\vec{PR} = \begin{pmatrix} -1 - 4 \\ 2 - 1 \end{pmatrix} PR⃗=(−51)\vec{PR} = \begin{pmatrix} -5 \\ 1 \end{pmatrix}

    The vector PR⃗\vec{PR} is (−51)\boxed{\begin{pmatrix} -5 \\ 1 \end{pmatrix}}.

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