Parametric equations of perpendicular lines
$begingroup$
I'm having problems with this:
Find the parametric equation of the line that passes through
the point $(-1, 4, 5)$ and is perpendicular to the line:
$$x = -2 + t$$
$$y = 1 - t$$
$$z = 1 + 2t$$
vectors analytic-geometry parametric
edited Jun 15 '15 at 20:54
Barry
2,094717
$endgroup$
add a comment |
$begingroup$
I'm having problems with this:
Find the parametric equation of the line that passes through
the point $(-1, 4, 5)$ and is perpendicular to the line:
$$x = -2 + t$$
$$y = 1 - t$$
$$z = 1 + 2t$$
vectors analytic-geometry parametric
edited Jun 15 '15 at 20:54
Barry
2,094717
$endgroup$
1
$begingroup$
wouldn't any line in the plane $x-y+2z = 5$ do?
$endgroup$
– abel
Jun 15 '15 at 20:59
add a comment |
$begingroup$
I'm having problems with this:
Find the parametric equation of the line that passes through
the point $(-1, 4, 5)$ and is perpendicular to the line:
$$x = -2 + t$$
$$y = 1 - t$$
$$z = 1 + 2t$$
vectors analytic-geometry parametric
edited Jun 15 '15 at 20:54
Barry
2,094717
$endgroup$
I'm having problems with this:
Find the parametric equation of the line that passes through
the point $(-1, 4, 5)$ and is perpendicular to the line:
$$x = -2 + t$$
$$y = 1 - t$$
$$z = 1 + 2t$$
vectors analytic-geometry parametric
vectors analytic-geometry parametric
edited Jun 15 '15 at 20:54
Barry
2,094717
edited Jun 15 '15 at 20:54
Barry
2,094717
edited Jun 15 '15 at 20:54
Barry
2,094717
edited Jun 15 '15 at 20:54
Barry
2,094717
edited Jun 15 '15 at 20:54
Barry
2,094717
2,094717
asked Jun 15 '15 at 20:48
user228932
1
$begingroup$
wouldn't any line in the plane $x-y+2z = 5$ do?
$endgroup$
– abel
Jun 15 '15 at 20:59
add a comment |
1
$begingroup$
wouldn't any line in the plane $x-y+2z = 5$ do?
$endgroup$
– abel
Jun 15 '15 at 20:59
1
1
$begingroup$
wouldn't any line in the plane $x-y+2z = 5$ do?
$endgroup$
– abel
Jun 15 '15 at 20:59
$begingroup$
wouldn't any line in the plane $x-y+2z = 5$ do?
$endgroup$
– abel
Jun 15 '15 at 20:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $X = (-2+t, 1-t, 1+2t)$ be a point on the line, and $x_0 = (-1,4,5)$ the given point. We want $(X - x_0) cdot (1, -1, 2) = 0$:
$$(-2 + t + 1)(1) + (1 - t - 4)(-1) + (1+2t -5)(2) = 0$$
$$6t - 6= 0$$
$$t = 1$$
Pick $x_1$ to be the point on $X$ at $t=1$, so $(-1, 0, 3)$.
Now all we need is an equation connecting $x_0$ and $x_1$. One such is:
$$begin{split}
x &= -1\
y &= -4t + 4\
z &= -2t + 5end{split}$$
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
$endgroup$
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%2f1326703%2fparametric-equations-of-perpendicular-lines%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
$begingroup$
Let $X = (-2+t, 1-t, 1+2t)$ be a point on the line, and $x_0 = (-1,4,5)$ the given point. We want $(X - x_0) cdot (1, -1, 2) = 0$:
$$(-2 + t + 1)(1) + (1 - t - 4)(-1) + (1+2t -5)(2) = 0$$
$$6t - 6= 0$$
$$t = 1$$
Pick $x_1$ to be the point on $X$ at $t=1$, so $(-1, 0, 3)$.
Now all we need is an equation connecting $x_0$ and $x_1$. One such is:
$$begin{split}
x &= -1\
y &= -4t + 4\
z &= -2t + 5end{split}$$
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
$endgroup$
add a comment |
$begingroup$
Let $X = (-2+t, 1-t, 1+2t)$ be a point on the line, and $x_0 = (-1,4,5)$ the given point. We want $(X - x_0) cdot (1, -1, 2) = 0$:
$$(-2 + t + 1)(1) + (1 - t - 4)(-1) + (1+2t -5)(2) = 0$$
$$6t - 6= 0$$
$$t = 1$$
Pick $x_1$ to be the point on $X$ at $t=1$, so $(-1, 0, 3)$.
Now all we need is an equation connecting $x_0$ and $x_1$. One such is:
$$begin{split}
x &= -1\
y &= -4t + 4\
z &= -2t + 5end{split}$$
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
$endgroup$
add a comment |
$begingroup$
Let $X = (-2+t, 1-t, 1+2t)$ be a point on the line, and $x_0 = (-1,4,5)$ the given point. We want $(X - x_0) cdot (1, -1, 2) = 0$:
$$(-2 + t + 1)(1) + (1 - t - 4)(-1) + (1+2t -5)(2) = 0$$
$$6t - 6= 0$$
$$t = 1$$
Pick $x_1$ to be the point on $X$ at $t=1$, so $(-1, 0, 3)$.
Now all we need is an equation connecting $x_0$ and $x_1$. One such is:
$$begin{split}
x &= -1\
y &= -4t + 4\
z &= -2t + 5end{split}$$
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
$endgroup$
Let $X = (-2+t, 1-t, 1+2t)$ be a point on the line, and $x_0 = (-1,4,5)$ the given point. We want $(X - x_0) cdot (1, -1, 2) = 0$:
$$(-2 + t + 1)(1) + (1 - t - 4)(-1) + (1+2t -5)(2) = 0$$
$$6t - 6= 0$$
$$t = 1$$
Pick $x_1$ to be the point on $X$ at $t=1$, so $(-1, 0, 3)$.
Now all we need is an equation connecting $x_0$ and $x_1$. One such is:
$$begin{split}
x &= -1\
y &= -4t + 4\
z &= -2t + 5end{split}$$
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
answered Jun 15 '15 at 21:06
BarryBarry
2,094717
2,094717
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%2f1326703%2fparametric-equations-of-perpendicular-lines%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
1
$begingroup$
wouldn't any line in the plane $x-y+2z = 5$ do?
$endgroup$
– abel
Jun 15 '15 at 20:59