Mixing
How is this algebra derived?
$begingroup$
In Gilbert Strang's linear algebra book, question 29 in section 1.2 gives:
$x + y + z = 0$
It then asks why, given that vector v = (x, y, z) and w = (z, x, y), that the value of $frac{vcdot w}{||v|| ||w||}$ is always $-frac12$.
I began solving it on my own but got stuck at xz + yz + xy. When I looked at the solution it gave me this:
$x⋅y = xz+yz+xy=frac 12(x+y+z)^2−frac12(x^2+y^2+z^2)$
My question is how is this done? How did they get the right hand side?
The solution ends by going from the algebra above to stating $vcdot w = 0 - frac 12 ||v||||w||.$ Then $
costheta = -frac 12$. This part I understand just fine.
Similar question was asked here: Converting dot producto to set of arithmetic mean differences?, and it gives me a clue with the arithmetic mean difference, but that I don't understand.
linear-algebra algebra-precalculus
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
$endgroup$
add a comment |
$begingroup$
In Gilbert Strang's linear algebra book, question 29 in section 1.2 gives:
$x + y + z = 0$
It then asks why, given that vector v = (x, y, z) and w = (z, x, y), that the value of $frac{vcdot w}{||v|| ||w||}$ is always $-frac12$.
I began solving it on my own but got stuck at xz + yz + xy. When I looked at the solution it gave me this:
$x⋅y = xz+yz+xy=frac 12(x+y+z)^2−frac12(x^2+y^2+z^2)$
My question is how is this done? How did they get the right hand side?
The solution ends by going from the algebra above to stating $vcdot w = 0 - frac 12 ||v||||w||.$ Then $
costheta = -frac 12$. This part I understand just fine.
Similar question was asked here: Converting dot producto to set of arithmetic mean differences?, and it gives me a clue with the arithmetic mean difference, but that I don't understand.
linear-algebra algebra-precalculus
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
$endgroup$
2
$begingroup$
$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2zy$
$endgroup$
– Vasya
May 15 '18 at 12:58
add a comment |
$begingroup$
In Gilbert Strang's linear algebra book, question 29 in section 1.2 gives:
$x + y + z = 0$
It then asks why, given that vector v = (x, y, z) and w = (z, x, y), that the value of $frac{vcdot w}{||v|| ||w||}$ is always $-frac12$.
I began solving it on my own but got stuck at xz + yz + xy. When I looked at the solution it gave me this:
$x⋅y = xz+yz+xy=frac 12(x+y+z)^2−frac12(x^2+y^2+z^2)$
My question is how is this done? How did they get the right hand side?
The solution ends by going from the algebra above to stating $vcdot w = 0 - frac 12 ||v||||w||.$ Then $
costheta = -frac 12$. This part I understand just fine.
Similar question was asked here: Converting dot producto to set of arithmetic mean differences?, and it gives me a clue with the arithmetic mean difference, but that I don't understand.
linear-algebra algebra-precalculus
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
$endgroup$
In Gilbert Strang's linear algebra book, question 29 in section 1.2 gives:
$x + y + z = 0$
It then asks why, given that vector v = (x, y, z) and w = (z, x, y), that the value of $frac{vcdot w}{||v|| ||w||}$ is always $-frac12$.
I began solving it on my own but got stuck at xz + yz + xy. When I looked at the solution it gave me this:
$x⋅y = xz+yz+xy=frac 12(x+y+z)^2−frac12(x^2+y^2+z^2)$
My question is how is this done? How did they get the right hand side?
The solution ends by going from the algebra above to stating $vcdot w = 0 - frac 12 ||v||||w||.$ Then $
costheta = -frac 12$. This part I understand just fine.
Similar question was asked here: Converting dot producto to set of arithmetic mean differences?, and it gives me a clue with the arithmetic mean difference, but that I don't understand.
linear-algebra algebra-precalculus
linear-algebra algebra-precalculus
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
asked May 15 '18 at 12:55
fuzzylogicalfuzzylogical
132
132
2
$begingroup$
$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2zy$
$endgroup$
– Vasya
May 15 '18 at 12:58
add a comment |
2
$begingroup$
$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2zy$
$endgroup$
– Vasya
May 15 '18 at 12:58
2
2
$begingroup$
$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2zy$
$endgroup$
– Vasya
May 15 '18 at 12:58
$begingroup$
$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2zy$
$endgroup$
– Vasya
May 15 '18 at 12:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Vasya's comment is correct. We start by expanding the square as follows...
$$(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$$
From this expression we can derive the formula you want in 3 steps.
Factor out the 2
$$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$$Subtract the squares
$$(x+y+z)^2 - (x^2+y^2+z^2) = 2(xy+yz+zx)$$Divide by 2
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx$$
Now we just notice that the RHS is the inner product of your two vectors $v$ and $w$ and we're done.
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx = vcdot w$$
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2782217%2fhow-is-this-algebra-derived%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Vasya's comment is correct. We start by expanding the square as follows...
$$(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$$
From this expression we can derive the formula you want in 3 steps.
Factor out the 2
$$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$$Subtract the squares
$$(x+y+z)^2 - (x^2+y^2+z^2) = 2(xy+yz+zx)$$Divide by 2
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx$$
Now we just notice that the RHS is the inner product of your two vectors $v$ and $w$ and we're done.
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx = vcdot w$$
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
$endgroup$
add a comment |
$begingroup$
Vasya's comment is correct. We start by expanding the square as follows...
$$(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$$
From this expression we can derive the formula you want in 3 steps.
Factor out the 2
$$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$$Subtract the squares
$$(x+y+z)^2 - (x^2+y^2+z^2) = 2(xy+yz+zx)$$Divide by 2
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx$$
Now we just notice that the RHS is the inner product of your two vectors $v$ and $w$ and we're done.
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx = vcdot w$$
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
$endgroup$
add a comment |
$begingroup$
Vasya's comment is correct. We start by expanding the square as follows...
$$(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$$
From this expression we can derive the formula you want in 3 steps.
Factor out the 2
$$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$$Subtract the squares
$$(x+y+z)^2 - (x^2+y^2+z^2) = 2(xy+yz+zx)$$Divide by 2
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx$$
Now we just notice that the RHS is the inner product of your two vectors $v$ and $w$ and we're done.
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx = vcdot w$$
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
$endgroup$
Vasya's comment is correct. We start by expanding the square as follows...
$$(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$$
From this expression we can derive the formula you want in 3 steps.
Factor out the 2
$$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$$Subtract the squares
$$(x+y+z)^2 - (x^2+y^2+z^2) = 2(xy+yz+zx)$$Divide by 2
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx$$
Now we just notice that the RHS is the inner product of your two vectors $v$ and $w$ and we're done.
$$frac{1}{2}(x+y+z)^2 - frac{1}{2}(x^2+y^2+z^2) = xy+yz+zx = vcdot w$$
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
edited Jan 14 at 20:12
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
answered May 15 '18 at 13:37
Nathan CrockNathan Crock
515
515
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2782217%2fhow-is-this-algebra-derived%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
$(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2zy$
$endgroup$
– Vasya
May 15 '18 at 12:58