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Chapter 14: Problem 6

Point P has coordinates \((7,8)\). Point \(Q\) has coordinates \((-2,4)\) (a) Draw a sketch showing \(\mathrm{P}\) and \(\mathrm{Q}\). (b) State the position vectors of \(\mathrm{P}\) and \(\mathrm{Q}\). (c) Find an expression for \(\overrightarrow{P Q}\). (d) Find \(|\overrightarrow{\mathrm{PQ}}|\).

Short Answer

Expert verified
Answer: The expression for the vector from point P to point Q is \(\overrightarrow{PQ} = \begin{pmatrix} -9 \\ -4 \end{pmatrix}\), and its magnitude is \(\sqrt{97}\).

Step by step solution

01

(a) Drawing a Sketch of P and Q)

To draw a sketch of points P and Q, plot the given coordinates (\((7,8)\) for point P and \((-2,4)\) for point Q) on the Cartesian plane.
02

(b) Position Vectors of P and Q)

To find the position vectors of points P and Q, write the given coordinates as column vectors: Position vector of P: \(\begin{pmatrix} 7 \\ 8 \end{pmatrix}\) Position vector of Q: \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\)
03

(c) Expression for \(\overrightarrow{PQ}\))

To find the expression for \(\overrightarrow{PQ}\), subtract the position vector of P from the position vector of Q, that is: \(\overrightarrow{PQ} = \begin{pmatrix} -2 \\ 4 \end{pmatrix} - \begin{pmatrix} 7 \\ 8 \end{pmatrix}\) Now perform the subtraction: \(\overrightarrow{PQ} = \begin{pmatrix} -2 - 7 \\ 4 - 8 \end{pmatrix} = \begin{pmatrix} -9 \\ -4 \end{pmatrix}\)
04

(d) Magnitude of \(\overrightarrow{PQ}\))

To find the magnitude of \(\overrightarrow{PQ}\), use the magnitude formula for a 2-dimensional vector: \(|\overrightarrow{PQ}| = \sqrt{(-9)^2 + (-4)^2}\) Calculate the magnitude: \(|\overrightarrow{PQ}| = \sqrt{81 + 16} = \sqrt{97}\) So, the magnitude of \(\overrightarrow{PQ}\) is \(\sqrt{97}\).

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Most popular questions from this chapter

Find the modulus of the vector $$ p=2 i-j+5 k $$

Explain the distinction between a position vector and a more general or free vector.

Write down the vector equation of the line passing through the points with position vectors $$ \begin{aligned} &p=3 i+7 j-2 k \\ &q=-3 i+2 j+2 k \end{aligned} $$ Find also the cartesian equation of this line.

If the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) is equal to zero, then (i) \(a=0\), or \(b=0\), or \(c=0\) or (ii) two of the vectors are parallel, or (iii) the three vectors lie in the same plane (they are said to be coplanar). Show that the vectors $$ 2 \boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}-4 \boldsymbol{j}+5 \boldsymbol{k}, \boldsymbol{i}+2 \boldsymbol{j}-3 \boldsymbol{k} $$ are coplanar.

State the coordinates of the point \(\mathrm{P}\) if its position vector is given as (a) \(3 \boldsymbol{i}-7 \boldsymbol{j}\), (b) \(-4 \boldsymbol{i}\), (c) \(-0.5 \boldsymbol{i}+13 \boldsymbol{j}\), (d) \(a \boldsymbol{i}+b \boldsymbol{j}\).
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