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Given k = 4 −3 and l = -5 7 .

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4 parts
a(i)
13.5
a(ii)

(−1417)\begin{pmatrix} -14 \\ 17 \end{pmatrix}

b(i)
5
5
b(ii)
-2

Step-by-step solution

13 steps

Here's the solution to the vector problem.

Part a(i)· 3 steps

Calculate ∣l−k∣|l - k|, correct to one decimal place.

Given k=(4−3)k = \begin{pmatrix} 4 \\ -3 \end{pmatrix} and l=(−57)l = \begin{pmatrix} -5 \\ 7 \end{pmatrix}.

  1. Calculate the vector l−kl - k.

    l−k=(−57)−(4−3)=(−5−47−(−3))=(−910)l - k = \begin{pmatrix} -5 \\ 7 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \end{pmatrix} = \begin{pmatrix} -5 - 4 \\ 7 - (-3) \end{pmatrix} = \begin{pmatrix} -9 \\ 10 \end{pmatrix}

  2. Calculate the magnitude of l−kl - k.

    ∣l−k∣=(−9)2+(10)2=81+100=181|l - k| = \sqrt{(-9)^2 + (10)^2} = \sqrt{81 + 100} = \sqrt{181}

  3. Round the magnitude to one decimal place.

    181≈13.4536...\sqrt{181} \approx 13.4536... ∣l−k∣≈13.5|l - k| \approx \boxed{13.5}

Part a(ii)

(−1417)\begin{pmatrix} -14 \\ 17 \end{pmatrix}

Find the vector rr such that k+r=2lk + r = 2l.

  1. Rearrange the equation to solve for rr.

    r=2l−kr = 2l - k

  2. Substitute the given vectors kk and ll.

    r=2(−57)−(4−3)r = 2 \begin{pmatrix} -5 \\ 7 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \end{pmatrix}

  3. Perform the scalar multiplication and vector subtraction.

    r=(2×−52×7)−(4−3)=(−1014)−(4−3)=(−10−414−(−3))=(−1417)r = \begin{pmatrix} 2 \times -5 \\ 2 \times 7 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \end{pmatrix} = \begin{pmatrix} -10 \\ 14 \end{pmatrix} - \begin{pmatrix} 4 \\ -3 \end{pmatrix} = \begin{pmatrix} -10 - 4 \\ 14 - (-3) \end{pmatrix} = \begin{pmatrix} -14 \\ 17 \end{pmatrix}

    The vector rr is (−1417)\boxed{\begin{pmatrix} -14 \\ 17 \end{pmatrix}}.

Part b(i)5

Find $|AB|$ and $|BA|$.

Given that A(1,3)A(1,3), B(−2,−1)B(-2,-1) and C(2,3n)C(2,3n) where nn is a constant.

  1. Calculate the vector AB⃗\vec{AB}.

    AB⃗=B−A=(−2−1−1−3)=(−3−4)\vec{AB} = B - A = \begin{pmatrix} -2 - 1 \\ -1 - 3 \end{pmatrix} = \begin{pmatrix} -3 \\ -4 \end{pmatrix}

  2. Calculate the magnitude $|AB|$.

    ∣AB∣=(−3)2+(−4)2=9+16=25=5|AB| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

    So, ∣AB∣=5|AB| = \boxed{5}.

  3. Calculate the vector BA⃗\vec{BA}.

    BA⃗=A−B=(1−(−2)3−(−1))=(1+23+1)=(34)\vec{BA} = A - B = \begin{pmatrix} 1 - (-2) \\ 3 - (-1) \end{pmatrix} = \begin{pmatrix} 1 + 2 \\ 3 + 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}

  4. Calculate the magnitude $|BA|$.

    ∣BA∣=(3)2+(4)2=9+16=25=5|BA| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

    So, ∣BA∣=5|BA| = \boxed{5}.

Part b(ii)-2

Find the value of nn if BC⃗=(4−5)\vec{BC} = \begin{pmatrix} 4 \\ -5 \end{pmatrix}.

  1. Express the vector BC⃗\vec{BC} using the coordinates of BB and CC.

    B(−2,−1)C(2,3n)B(-2,-1) \quad C(2,3n) BC⃗=C−B=(2−(−2)3n−(−1))=(2+23n+1)=(43n+1)\vec{BC} = C - B = \begin{pmatrix} 2 - (-2) \\ 3n - (-1) \end{pmatrix} = \begin{pmatrix} 2 + 2 \\ 3n + 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 3n + 1 \end{pmatrix}

  2. Equate this expression for BC⃗\vec{BC} to the given vector (4−5)\begin{pmatrix} 4 \\ -5 \end{pmatrix}.

    (43n+1)=(4−5)\begin{pmatrix} 4 \\ 3n + 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \end{pmatrix}

  3. Equate the y-components to solve for nn.

    3n+1=−53n + 1 = -5 3n=−5−13n = -5 - 1 3n=−63n = -6 n=−63n = \frac{-6}{3} n=−2n = \boxed{-2}
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