Question

Let v = (7, 8). Suppose w ∈ double-struck R2 is perpendicular to v, and that w = 7. This determines w up to sign. Find one such w

Answer 1

Answer:

As per the given statement:

Let w ∈$R^2$ i.e w = (a, b)

then,

by perpendicular to v means;

$w \cdot v=0$

$(a, b) \cdot (7, 8)=0$

$7a + 8b = 0$

or

$7a= -8b$

Divide both sides by 7 we get;

$a = -\frac{8}{7}b$                       ......[1]

so, $w = (-\frac{8}{7}b, b)$ for some values of b.

Using the fact: ||w|| = 7 we get;

$||w|| = \sqrt{(-\frac{8}{7}b)^2+b^2 }$

substitute the given values, we have;

$7= \sqrt{(-\frac{8}{7}b)^2+b^2}$

Squaring both sides we have;

$7^2 =\frac{64}{49}b^2+b^2$

Simplify:

$49 = \frac{113}{49}b^2$

Or

$b^2 = 49 \times \frac{49}{113}$

or

$b = \pm \sqrt{49 \times \frac{49}{113}}=7 \cdot \frac{7}{\sqrt{113}} =\pm\frac{49}{\sqrt{113}}$

Substitute the value of b in [1] we get;

$a =- \frac{8}{7}b =-\frac{8}{7} \cdot \frac{49}{\sqrt{113}} = - \frac{56}{\sqrt{113}}$

therefore, two such answers of w are;

$w = \pm (-\frac{56}{\sqrt{113}}, \frac{49}{\sqrt{113}})$