The vectors 6i + 8j and 8i - 6j are parallel to →OP and →OQ...
The vectors 6i + 8j and 8i - 6j are parallel to →OP and →OQ respectively. If the magnitude of →OP and →OQ are 80 units and 120 units respectively, express: →OP and →OQ in terms of i and j;
ii. |→PQ|, in the form c√k, where c and k are constants.
Explanation
The gradient of two parallel lines are equal.
m1 = m2
6i + 8j is parallel to →OP
m = \(\frac{y_2 - y_1}{x_2 - x_1}\)
gradient(m) of 6i + 8j
m = \(\frac{8}{6}\) = \(\frac{4}{3}\)
\(\frac{y}{x}\) = \(\frac{4}{3}\)
3y - 4x =0
|OP| = 80
|OP| = √(x\(^2\) + y\(^2\)) = 80
x\(^2\) + y\(^2\) = 80\(^2\)
x\(^2\) + y\(^2\) = 6400
\(y = \frac{4x}{3}\)
\(x^2 + ({\frac{4x}{3}})^2\) = 6400
\(x^2 + \frac{16x^2}{9}\) = 6400
multiply through by 9
9x\(^2\) + 16x2 = 57600
25x\(^2\) = 57600
x\(^2\) = \(\frac{57600}{25}\)
x\(^2\) = 2304
x = √2304
x = 48
\(y = \frac{4x}{3}\) → \(y = \frac{4*48}{3}\)
y = 64
→OP =xi + yj
→OP = 48i + 64j
8i - 6j is parallel to →OQ
gradient of 8i - 6j = \(\frac{-6}{8}\)
\(\frac{y}{x}\) = \(\frac{-6}{8}\)
y = \(\frac{-6x}{8}\)
\(x^2 + ({\frac{-6x}{8}})^2\) = 14400
\(x^2 + \frac{36x^2}{64}\) = 14400
multiply through by 64
64x\(^2\) + 36x\(^2\) = 921600
100x\(^2\) = 921600
divide both sides by 100
x\(^2\) = 9216
find the square root of both sides
x = √9216
x = 96
y = \(\frac{-6x}{8}\)
y = \(\frac{-6[96]}{8}\)
y = -72
→OQ = xi + yj
→OQ = 96i - 72j
→PQ = (96i - 72j) - (48i + 64j)
→PQ = (96-48)i + (-72-64)j
→PQ = 48i - 136j
|→PQ| = √(48\(^2\) + (-136)\(^2\))
|→PQ| = √(2304 + 18496)
|→PQ| = √20800
|→PQ| = 40√13
c = 40, k = 13