Mixing
Foot of perpendicular in 3 dimensions
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Let us suppose that $Q$ is a foot of perpendicular from a point $P (2,4,3) $on the line joining the points $A(1,2,4)$ and $B(3,4,5)$; then what are the coordinates of $Q$?
My try: Had this been a two dimensional problem, I would have found the equation of line and proceeded further to get the answer. But in 3D neither do I know to find the equation of a line nor do I know how to proceed further. Kindly Help.
3d
asked Feb 2 '17 at 17:14
Zlatan
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bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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Let us suppose that $Q$ is a foot of perpendicular from a point $P (2,4,3) $on the line joining the points $A(1,2,4)$ and $B(3,4,5)$; then what are the coordinates of $Q$?
My try: Had this been a two dimensional problem, I would have found the equation of line and proceeded further to get the answer. But in 3D neither do I know to find the equation of a line nor do I know how to proceed further. Kindly Help.
3d
asked Feb 2 '17 at 17:14
Zlatan
369212
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Are you familiar with vector calculus, or at least linear algebra?
– Triatticus
Feb 2 '17 at 17:17
I am. I just need to brush up the concepts, so if you could give me a clue how to proceed?
– Zlatan
Feb 2 '17 at 17:20
You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points.
– Triatticus
Feb 2 '17 at 17:22
Could you start with the solution please?
– Zlatan
Feb 2 '17 at 17:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let us suppose that $Q$ is a foot of perpendicular from a point $P (2,4,3) $on the line joining the points $A(1,2,4)$ and $B(3,4,5)$; then what are the coordinates of $Q$?
My try: Had this been a two dimensional problem, I would have found the equation of line and proceeded further to get the answer. But in 3D neither do I know to find the equation of a line nor do I know how to proceed further. Kindly Help.
3d
asked Feb 2 '17 at 17:14
Zlatan
369212
Let us suppose that $Q$ is a foot of perpendicular from a point $P (2,4,3) $on the line joining the points $A(1,2,4)$ and $B(3,4,5)$; then what are the coordinates of $Q$?
My try: Had this been a two dimensional problem, I would have found the equation of line and proceeded further to get the answer. But in 3D neither do I know to find the equation of a line nor do I know how to proceed further. Kindly Help.
3d
3d
asked Feb 2 '17 at 17:14
Zlatan
369212
asked Feb 2 '17 at 17:14
Zlatan
369212
asked Feb 2 '17 at 17:14
Zlatan
369212
asked Feb 2 '17 at 17:14
Zlatan
369212
asked Feb 2 '17 at 17:14
Zlatan
369212
369212
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Are you familiar with vector calculus, or at least linear algebra?
– Triatticus
Feb 2 '17 at 17:17
I am. I just need to brush up the concepts, so if you could give me a clue how to proceed?
– Zlatan
Feb 2 '17 at 17:20
You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points.
– Triatticus
Feb 2 '17 at 17:22
Could you start with the solution please?
– Zlatan
Feb 2 '17 at 17:28
add a comment |
Are you familiar with vector calculus, or at least linear algebra?
– Triatticus
Feb 2 '17 at 17:17
I am. I just need to brush up the concepts, so if you could give me a clue how to proceed?
– Zlatan
Feb 2 '17 at 17:20
You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points.
– Triatticus
Feb 2 '17 at 17:22
Could you start with the solution please?
– Zlatan
Feb 2 '17 at 17:28
Are you familiar with vector calculus, or at least linear algebra?
– Triatticus
Feb 2 '17 at 17:17
Are you familiar with vector calculus, or at least linear algebra?
– Triatticus
Feb 2 '17 at 17:17
I am. I just need to brush up the concepts, so if you could give me a clue how to proceed?
– Zlatan
Feb 2 '17 at 17:20
I am. I just need to brush up the concepts, so if you could give me a clue how to proceed?
– Zlatan
Feb 2 '17 at 17:20
You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points.
– Triatticus
Feb 2 '17 at 17:22
You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points.
– Triatticus
Feb 2 '17 at 17:22
Could you start with the solution please?
– Zlatan
Feb 2 '17 at 17:28
Could you start with the solution please?
– Zlatan
Feb 2 '17 at 17:28
add a comment |
1 Answer
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We know $A(1,2,4)$ and $B(3,4,5)$, so $vec{AB}=langle2,2,1rangle$.
A vector equation for $overleftrightarrow{AB}$ is $vec{r}=langle1,2,4rangle+tlangle2,2,1rangle$ for $tinmathbb{R}$.
Point $Q$ must lie on $overleftrightarrow{AB}$, so the position vector for $Q$ is $vec{Q}=langle1,2,4rangle+qlangle2,2,1rangle=langle1+2q,2+2q,4+qrangle$ for some $qinmathbb{R}$.
We need to find $q$ in order to find $Q$. We know $Q$ is the foot of the perpendicular from $P$ to $Q$. In other words, $vec{PQ}perpvec{AB}$.
We find $vec{PQ}=langle2q-1,2q-2,q+1rangle$.
Since $vec{PQ}perpvec{AB}$, then $vec{PQ}cdotvec{AB}=0$.
begin{align}
vec{PQ}cdotvec{AB}&=0\
langle2q-1,2q-2,q+1ranglecdotlangle2,2,1rangle&=0\
4q-2+4q-4+q+1&=0\
9q&=5\
q&=frac{5}{9}
end{align}
Using $q=5/9$, we find that $vec{Q}=langle1+2(5/9),2+2(5/9),4+(5/9)rangle=langle19/9,28/9,41/9rangle$.
Thus we have $Qleft(frac{19}{9},frac{28}{9},frac{41}{9}right)$.
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
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1 Answer
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1 Answer
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We know $A(1,2,4)$ and $B(3,4,5)$, so $vec{AB}=langle2,2,1rangle$.
A vector equation for $overleftrightarrow{AB}$ is $vec{r}=langle1,2,4rangle+tlangle2,2,1rangle$ for $tinmathbb{R}$.
Point $Q$ must lie on $overleftrightarrow{AB}$, so the position vector for $Q$ is $vec{Q}=langle1,2,4rangle+qlangle2,2,1rangle=langle1+2q,2+2q,4+qrangle$ for some $qinmathbb{R}$.
We need to find $q$ in order to find $Q$. We know $Q$ is the foot of the perpendicular from $P$ to $Q$. In other words, $vec{PQ}perpvec{AB}$.
We find $vec{PQ}=langle2q-1,2q-2,q+1rangle$.
Since $vec{PQ}perpvec{AB}$, then $vec{PQ}cdotvec{AB}=0$.
begin{align}
vec{PQ}cdotvec{AB}&=0\
langle2q-1,2q-2,q+1ranglecdotlangle2,2,1rangle&=0\
4q-2+4q-4+q+1&=0\
9q&=5\
q&=frac{5}{9}
end{align}
Using $q=5/9$, we find that $vec{Q}=langle1+2(5/9),2+2(5/9),4+(5/9)rangle=langle19/9,28/9,41/9rangle$.
Thus we have $Qleft(frac{19}{9},frac{28}{9},frac{41}{9}right)$.
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
add a comment |
up vote
0
down vote
We know $A(1,2,4)$ and $B(3,4,5)$, so $vec{AB}=langle2,2,1rangle$.
A vector equation for $overleftrightarrow{AB}$ is $vec{r}=langle1,2,4rangle+tlangle2,2,1rangle$ for $tinmathbb{R}$.
Point $Q$ must lie on $overleftrightarrow{AB}$, so the position vector for $Q$ is $vec{Q}=langle1,2,4rangle+qlangle2,2,1rangle=langle1+2q,2+2q,4+qrangle$ for some $qinmathbb{R}$.
We need to find $q$ in order to find $Q$. We know $Q$ is the foot of the perpendicular from $P$ to $Q$. In other words, $vec{PQ}perpvec{AB}$.
We find $vec{PQ}=langle2q-1,2q-2,q+1rangle$.
Since $vec{PQ}perpvec{AB}$, then $vec{PQ}cdotvec{AB}=0$.
begin{align}
vec{PQ}cdotvec{AB}&=0\
langle2q-1,2q-2,q+1ranglecdotlangle2,2,1rangle&=0\
4q-2+4q-4+q+1&=0\
9q&=5\
q&=frac{5}{9}
end{align}
Using $q=5/9$, we find that $vec{Q}=langle1+2(5/9),2+2(5/9),4+(5/9)rangle=langle19/9,28/9,41/9rangle$.
Thus we have $Qleft(frac{19}{9},frac{28}{9},frac{41}{9}right)$.
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
add a comment |
up vote
0
down vote
up vote
0
down vote
We know $A(1,2,4)$ and $B(3,4,5)$, so $vec{AB}=langle2,2,1rangle$.
A vector equation for $overleftrightarrow{AB}$ is $vec{r}=langle1,2,4rangle+tlangle2,2,1rangle$ for $tinmathbb{R}$.
Point $Q$ must lie on $overleftrightarrow{AB}$, so the position vector for $Q$ is $vec{Q}=langle1,2,4rangle+qlangle2,2,1rangle=langle1+2q,2+2q,4+qrangle$ for some $qinmathbb{R}$.
We need to find $q$ in order to find $Q$. We know $Q$ is the foot of the perpendicular from $P$ to $Q$. In other words, $vec{PQ}perpvec{AB}$.
We find $vec{PQ}=langle2q-1,2q-2,q+1rangle$.
Since $vec{PQ}perpvec{AB}$, then $vec{PQ}cdotvec{AB}=0$.
begin{align}
vec{PQ}cdotvec{AB}&=0\
langle2q-1,2q-2,q+1ranglecdotlangle2,2,1rangle&=0\
4q-2+4q-4+q+1&=0\
9q&=5\
q&=frac{5}{9}
end{align}
Using $q=5/9$, we find that $vec{Q}=langle1+2(5/9),2+2(5/9),4+(5/9)rangle=langle19/9,28/9,41/9rangle$.
Thus we have $Qleft(frac{19}{9},frac{28}{9},frac{41}{9}right)$.
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
We know $A(1,2,4)$ and $B(3,4,5)$, so $vec{AB}=langle2,2,1rangle$.
A vector equation for $overleftrightarrow{AB}$ is $vec{r}=langle1,2,4rangle+tlangle2,2,1rangle$ for $tinmathbb{R}$.
Point $Q$ must lie on $overleftrightarrow{AB}$, so the position vector for $Q$ is $vec{Q}=langle1,2,4rangle+qlangle2,2,1rangle=langle1+2q,2+2q,4+qrangle$ for some $qinmathbb{R}$.
We need to find $q$ in order to find $Q$. We know $Q$ is the foot of the perpendicular from $P$ to $Q$. In other words, $vec{PQ}perpvec{AB}$.
We find $vec{PQ}=langle2q-1,2q-2,q+1rangle$.
Since $vec{PQ}perpvec{AB}$, then $vec{PQ}cdotvec{AB}=0$.
begin{align}
vec{PQ}cdotvec{AB}&=0\
langle2q-1,2q-2,q+1ranglecdotlangle2,2,1rangle&=0\
4q-2+4q-4+q+1&=0\
9q&=5\
q&=frac{5}{9}
end{align}
Using $q=5/9$, we find that $vec{Q}=langle1+2(5/9),2+2(5/9),4+(5/9)rangle=langle19/9,28/9,41/9rangle$.
Thus we have $Qleft(frac{19}{9},frac{28}{9},frac{41}{9}right)$.
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
edited Feb 2 '17 at 18:42
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
answered Feb 2 '17 at 18:23
Tim Thayer
1,186410
1,186410
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Are you familiar with vector calculus, or at least linear algebra?
– Triatticus
Feb 2 '17 at 17:17
I am. I just need to brush up the concepts, so if you could give me a clue how to proceed?
– Zlatan
Feb 2 '17 at 17:20
You are trying to find the component of the vector that ends on the point in question along the line defined by the other two points.
– Triatticus
Feb 2 '17 at 17:22
Could you start with the solution please?
– Zlatan
Feb 2 '17 at 17:28