The position vectors of P, Q and R with respect to the origin are (4i-5j), (i+3j) and (-5i+2j) respectively. If PQRM is a parallelogram, find: (a) the coord...

Answer Details

To solve this problem, we first need to understand what a parallelogram is. A parallelogram is a four-sided shape with opposite sides that are parallel to each other. This means that if we draw lines from opposite corners of the parallelogram, these lines will intersect at the midpoint of the two sides they are drawn from.

Now let's look at the given position vectors of P, Q, and R. We can use these to find the position vector of M, which is opposite to P in the parallelogram. To do this, we need to use the fact that the position vector of the midpoint of a line segment is the average of the position vectors of the two endpoints. So, the position vector of M is:

¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ =  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ - (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ = (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

MR¯¯¯¯¯¯¯¯¯ $\overline{𝑀𝑅}$ =  (−52) $\left(\begin{array}{c}-5\\ 2\end{array}\right)$ - (xy) $\left(\begin{array}{c}𝑥\\ 𝑦\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

 (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

⇒ -5 - x = -3; x = -2

⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)

(b) PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$ = OM¯¯¯¯¯¯¯¯¯ $\overline{𝑂𝑀}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=   (−2−6) $\left(\begin{array}{c}-2\\ -6\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−6−1) $\left(\begin{array}{c}-6\\ -1\end{array}\right)$

    PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ = OQ¯¯¯¯¯¯¯¯ $\overline{𝑂𝑄}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

|PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$| = √(-62 ${}^2$ + [-12 ${}^2$]) = √37

|PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$| = √(-32 ${}^2$ + 82 ${}^2$) = √73

cosØ = PM.PQ|PM¯¯¯¯¯¯¯¯¯¯||PQ¯¯¯¯¯¯¯¯| $\frac{𝑃𝑀.𝑃𝑄}{|\overline{𝑃𝑀}||\overline{𝑃𝑄}|}$

cosØ = [−6i−j].[−3i+8j]√37.√73 $\frac{[-6𝑖-𝑗].[-3𝑖+8𝑗]}{\surd 37.\surd 73}$ = 10√2710 $\frac{10}{\surd 2710}$

⇒ Ø = cos−1 ${}^{-1}$ 10√2710 $\frac{10}{\surd 2710}$

: Ø = 78.91° ≈ 79°