# Finding Possible Values for K in Vector Algebra

• Accurim
In summary, Daniel says that there are multiple ways to solve for K, and that by squaring both sides of the equation, one of the answers will be correct.
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)$$\bullet (-1, k, 1) = \sqrt14\sqrt{2+k^2}\frac{3}{7}$$
$$7k-7=\sqrt14\sqrt{2+k^2}$$

Get all the k terms together on 1 side first.

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

Daniel.

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

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?

Last edited by a moderator:
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.

KoGs said:
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
$$\frac{9k^2}{k^2+2} = \frac{9(14)}{7} + \frac{9\sqrt14}{7\sqrt{k^2+2}} + \frac{9\sqrt14}{7\sqrt{k^2+2}} + \frac{9}{k^2+2}$$
Code:

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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.

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.

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

## What is the dot product in algebra?

The dot product in algebra is a mathematical operation that takes two vectors and produces a single scalar value. It is also known as the scalar product or inner product.

## How is the dot product calculated?

The dot product is calculated by multiplying the corresponding entries in two vectors and then adding the products together. For example, if vector A is (a1, a2, a3) and vector B is (b1, b2, b3), the dot product would be a1b1 + a2b2 + a3b3.

## What is the significance of the dot product?

The dot product is significant because it can be used to calculate the angle between two vectors, which can provide information about the relationship between the vectors. It is also used in various applications such as physics, engineering, and computer graphics.

## What are some properties of the dot product?

Some properties of the dot product include commutativity, distributivity, and associativity. It is also used to calculate the length (or magnitude) of a vector and can be used to determine if two vectors are perpendicular.

## In what situations is the dot product useful?

The dot product is useful in situations where we want to find the angle between two vectors or calculate the projection of one vector onto another. It is also used in optimization problems and in calculating work done by a force in physics.

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