Finding Possible Values for K in Vector Algebra

[Accurim]

Hi I'm new here and I'm having trouble with this algebra question please help. Sorry if my latex is ugly I'm new to it. I get stuck at the bottom line and I'm not sure how to go further with the question to solve for K

#7 - Angle between 2 vectors is $$\alpha$$ where cos$$\alpha$$ = $$\frac{3}{7}$$. a = (2,3,-1) and b = ( -1, K, 1) use the 2 vectors and find possible values for K.

This is what I did:
$$a\bullet c = |a||b|cos\alpha$$
(2,3,-1)$$\bullet (-1, k, 1) = \sqrt14\sqrt{2+k^2}\frac{3}{7}$$
$$7k-7=\sqrt14\sqrt{2+k^2}$$

 

[KoGs]

Get all the k terms together on 1 side first.

 

[dextercioby]

IF the eq you got is correct, then you can square it already in that form.

Daniel.

 

[KoGs]

That would make things measier than need be, because of the -7.

 

[Accurim]

I actually squared both sides and depending on when you square both sides you'll get different answers. I got 4 answers depending on when I square both sides. I tested the values of K and only 1 out of 4 them are correct, I'll show you what I did.

$$a\bullet b = |a||b|cos\alpha$$
$$(2,3,-1)\bullet(-1, K, 1) = (\sqrt{2^2+3^+{-1}^2})(\sqrt{k^2+(-1)^2+1^2})(\frac{3}{7})$$
$$-2 +3k -1 = \sqrt 14\sqrt{k^2+2}(\frac {3}{7})$$
$$3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})$$

Now you can simplify but multiplying both sides by 7 and dividing by 3 and it will create the line I wrote in the first post
$$7k-7=\sqrt 14\sqrt {k^2+2}$$
After squaring both sides and solving for K, i got k=-1/5 and =3, both values were incorrect when plugged back into the original equation.

However if you square both sides on this line one of the answers will be correct:
$$3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})$$
$$(3k-3)^2=(\sqrt 14)^2 (\sqrt {k^2+2})^2(\frac {3^2}{7^2})$$
$$9k^2 -18k + 9 = 14 (k^2+2)(\frac {9}{49})$$
$$9k^2 -18k + 9 = 2 (k^2+2)(\frac {9}{7})$$
$$9k^2 -18k + 9 = \frac {18}{7}(k^2+2)$$
$$63k^2 -126k + 63 = 18(k^2+2)$$
$$63k^2 -126k + 63 = 18k^2 + 36$$
$$45k^2 - 126k + 27 = 0$$
$$9 (5k^2-14k + 3) = 0$$
$$k = \frac {14 +- \sqrt 136}{10}$$
$$K = 2.5662 or k = 1.08377$$

Only K = 2.5662 actually worked when I tested it:
$$-2 +3k -1 = \sqrt 14\sqrt{k^2+2}(\frac {3}{7})$$
$$\frac {3k-3}{\sqrt 14\sqrt{k^2+2}}=\frac {3}{7}$$
Pluggin in K = 2.5662
$$0.4286= \frac {3}{7}$$http://www.jimloy.com/algebra/square.htm -> A link saying squaring both sides can give wrong answers.

So the question is asking for possible values for K, is there anyway I can solve for more?

 

[KoGs]

Show your work for this too: After squaring both sides and solving for K, i got k=-1/5 and =3, both values were incorrect when plugged back into the original equation.

Try the way I suggested, gather all the K's one one side. And everything else on the other side. See what happens.

 

[Accurim]

Show your work for this too: After squaring both sides and solving for K, i got k=-1/5 and =3, both values were incorrect when plugged back into the original equation.

Try the way I suggested, gather all the K's one one side. And everything else on the other side. See what happens.

For k=-1/5 and k=3

$$a\bullet b = |a||b|cos\alpha$$
$$(2,3,-1)\bullet(-1, K, 1) = (\sqrt{2^2+3^+{-1}^2})(\sqrt{k^2+(-1)^2+1^2})(\frac{3}{7})$$
$$-2 +3k -1 = \sqrt 14\sqrt{k^2+2}(\frac {3}{7})$$
$$3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})$$
$$7k-7=\sqrt 14\sqrt {k^2+2}$$
$$7(k-1)=\sqrt 14\sqrt {k^2+2}$$
$$7^2(k-1)^2=(\sqrt 14)^2(\sqrt {k^2+2})^2$$
$$49(k^2-2k+1) = 14k^2 + 28$$
$$49k^2-98k+49=14k^2+28$$
$$35k^2-98k+21=0$$
$$7(5k^2-14k+3)=0$$
$$7(5k+1)(k-3)=0$$

K= -1/5 or K=3


Okay it turns out being really messy so I didn't complete it.

$$3k-3=\sqrt14 \sqrt {k^2+2}(\frac {3}{7})$$
$$\frac{3k}{\sqrt{k^2+2}} = \frac {3\sqrt14}{7} + \frac{3}{\sqrt{k^2+2}}$$
Squaring both sides
$$\frac{9k^2}{k^2+2} = \frac{9(14)}{7} + \frac{9\sqrt14}{7\sqrt{k^2+2}} + \frac{9\sqrt14}{7\sqrt{k^2+2}} + \frac{9}{k^2+2}$$

 

[dextercioby]

The last line factorization is not correct. So your k=-1/5 and k=3 are not correct. By both methods you'll get the same result, as expected.

Daniel.

 

[KoGs]

Why did you backtrack a line? You should have squared it after you multipled through by 7/3, as you did for the other method(s).

Anyways yeah dextercioby is right. I didn't look too closely. Look closely at your signs. To test it, multiply your factorization through and see what you originally got before you started factorizing.

 

[Accurim]

Ah I missed that integer sign on the 3. Anyways I went with the answer I got before so thanks for the help guys.