Question

the angle between two vectors 4i+7j+6k and 3i+3j-(c/10)k is 60 degree. find 3c/17

Answer 1

Answer:

angle between two angle between two vectors 4i+7j+6k and angle between two vectors 4i+7j+6k and 3i+3j-(c/10)k is 60 degree is 60 degree 4i+7j+6k and 3i+3j-(c/10)k is 60 degree

Answer 2

Answer:

Required value of 3c/17 is $\frac{14640± 180 \sqrt{87610586} }{493}$

Explanation:

Given two vectors are

$4i + 7j + 6k$and $3i + 3j - \frac{c}{10} k$

Let $a = 4i + 7j + 6k$and $b = 3i + 3j - \frac{c}{10} k$

Now,$|a| = \sqrt{ {4}^{2} + {7}^{2} + {6}^{2} } = \sqrt{16 + 49 + 36} = \sqrt{101}$

and $|b| = \sqrt{ {3}^{2} + {3}^{2} + {( \frac{c}{10}) }^{2} } = \sqrt{18 + \frac{ {c}^{2} }{100} }$

Now $a.b = 12 + 27 - \frac{3c}{5} = 39 - \frac{3c}{5}$

As we know that,

$a.b = |a| |b| \cos(\theta)$

Here $\theta$is the angle between vectors a and b.

Therefore $39 - \frac{3c}{5} = \sqrt{101} \times \sqrt{18+ \frac{ {c}^{2} }{100} \cos( {60}^{o} ) }$

$1521 - \frac{234}{5} c + \frac{9 {c}^{2} }{25} = \frac{1}{2} ( 1818 + \frac{101 {c}^{2} }{100} ) \\ 3042 - \frac{468}{5}c + \frac{18 {c}^{2} }{25} = 1818+ \frac{101 {c}^{2} }{100}$

$-122400 + 29 {c}^{2} - 9360c = 0$

We are comparing this equation with

$p {x}^{2} + qx + r = 0$

So,p=29,q=-9360 and r=-122400

By Sridhar Acharya's law,

$x = \frac{9360± \sqrt{ {9360}^{2} + 4 \times 122400 \times 29} }{2 \times 29} = \frac{9360± \sqrt{101808000} }{58} = \frac{9360± 120 \sqrt{87610586} }{58} = \frac{4880± 60 \sqrt{87610586} }{29}$

Now $\frac{3c}{17} = \frac{3}{17} \times \frac{4880± 60 \sqrt{87610586} }{29} = \frac{14640± 180 \sqrt{87610586} }{493}$