find eqn of line thru (1,2,3), perpendicular to (3-2t,5+8t,7-4t) and (-2t, 5+t, 7-t)

[frank789]

hi all!

was doing this in homework

find the equation line through (1,2,3) and perpendicular to (3-2t,5+8t,7-4t) and (-2t, 5+t, 7-t).

thought it would be easy enough. crossed the two vectors to get a vector parallel to (1,2,3) however when i computed the cross product
i got quadratic equations.

the coefficient of the t^2 term in each quadratic given the correct t(a , b , c) to write my equation in the form of r(t) = (1,2,3) + t(a,b,c)

i dont understand what i did wrong here because it seems awfully incorrect to just use one coefficient and ignore the rest of the equation.

any help appreciated!

 

[pka]

was doing this in homework
find the equation line through (1,2,3) and perpendicular to (3-2t,5+8t,7-4t) and (-2t, 5+t, 7-t).
thought it would be easy enough. crossed the two vectors to get a vector parallel to (1,2,3).

Please clarify this question.
\(\displaystyle P: (1,2,3)\) is a point and each of \(\displaystyle (3-2t,5+8t,7-4t)~\&~ (-2t, 5+t, 7-t)\) is a line.

If that is correct then are they skew lines? If so there is a unique line that in perpendicular to each.
That line may not contain P.

Tells us if you mean something else.

 

[frank789]

(1,2,3) is a point on a line that is perpendicular to those two lines

they cant be skew lines because if they are both perpendicular to a certain point they must form a plane and therefore intersect? (thinking out loud there sorry)

 

[HallsofIvy]

You got a quadratic equation in what variable? The line (3- 2t, 5+ 8t, 7- 4t) had "direction vector" <-2, 8, -4>. The line (-2t, 5+ t, 7- t) has "direction vector" <-2, 1, -1>. A vector perpendicular to both is given by their cross product, \(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -2 & 8 & -4 \\ -2 & 1 & -1 \end{array}\right|\)= \(\displaystyle (-8+ 4)\vec{i}- (2- 8)\vec{j}+ (-2+ 16)\vec{k}\)\(\displaystyle = -4\vec{i}+ 6\vec{j}+ 14\vec{k}\) (check my arithmetic on this- I did it hastily).

So the line you want is the one through (1, 2, 3) with direction vector <-4, 6, 14>.

 

[frank789]

my mistake i should have just crossed the constant terms...thanks anyway!

 

[frank789]

You got a quadratic equation in what variable? The line (3- 2t, 5+ 8t, 7- 4t) had "direction vector" <-2, 8, -4>. The line (-2t, 5+ t, 7- t) has "direction vector" <-2, 1, -1>. A vector perpendicular to both is given by their cross product, \(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -2 & 8 & -4 \\ -2 & 1 & -1 \end{array}\right|= (-8+ 4)\vec{i}- (2- 8)\vec{j}+ (-2+ 16}\vec{k}= -4\vec{i}+ 6\vec{j}+ 14\vec{k}\) (check my arithmetic on this- I did it hastily).

So the line you want is the one through (1, 2, 3) with direction vector <-4, 6, 14>.

yes this sorry...long summer

 

[pka]

(1,2,3) is a point on a line that is perpendicular to those two lines
they cant be skew lines because if they are both perpendicular to a certain point they must form a plane and therefore intersect? (thinking out loud there sorry)

So each of \(\displaystyle (3-2t,5+8t,7-4t)~\&~ (-2s, 5+s, 7-s)\) is a line. Note that good form demands that you use different parameters.
If you think they intersect, you must find a valued of \(\displaystyle t\) and a value of \(\displaystyle s\) giving the same point. Can you?

 

[pka]

You got a quadratic equation in what variable? The line (3- 2t, 5+ 8t, 7- 4t) had "direction vector" <-2, 8, -4>. The line (-2t, 5+ t, 7- t) has "direction vector" <-2, 1, -1>. A vector perpendicular to both is given by their cross product, \(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -2 & 8 & -4 \\ -2 & 1 & -1 \end{array}\right|\)= \(\displaystyle (-8+ 4)\vec{i}- (2- 8)\vec{j}+ (-2+ 16)\vec{k}\)\(\displaystyle = -4\vec{i}+ 6\vec{j}+ 14\vec{k}\) (check my arithmetic on this- I did it hastily).
So the line you want is the one through (1, 2, 3) with direction vector <-4, 6, 14>.

These are two skew lines are they not?

 

[frank789]

So each of \(\displaystyle (3-2t,5+8t,7-4t)~\&~ (-2s, 5+s, 7-s)\) is a line. Note that good form demands that you use different parameters.
If you think they intersect, you must find a valued of \(\displaystyle t\) and a value of \(\displaystyle s\) giving the same point. Can you?

i ended up just crossing the direction vectors and getting the correct answer

to follow up with what your saying, do no think that such a value of t and exist

 

[pka]

i ended up just crossing the direction vectors and getting the correct answer
to follow up with what your saying, do no think that such a value of t and exist

Here are two methods: method I & method II .
As anyone can see there is no solution either way.
That means the two lines are skew lines since the are not parallel.

Now you say you got the correct answer. Have you checked that the answer intersects both lines.
(You can use WolfFramAlpha to check.)

 

[pka]

i ended up just crossing the direction vectors and getting the correct answer
to follow up with what your saying, do no think that such a value of t and exist

Using \(\displaystyle P: (1,2,3)\) is a point and each of \(\displaystyle (3-2t,5+8t,7-4t)~\&~ (-2s, 5+s, 7-s)\) is a line.
The vector \(\displaystyle <2,-3,-7>\) is perpendicular to each of those lines.

So the line \(\displaystyle \ell(u)=(1+2u,2-3u,3-7u)\) is the one you claim to be the correct answer.
However, look here & look here . You see that neither of the given lines intersects the line you claim to be correct.
If they do not intersect then they are not perpendicular.

Let \(\displaystyle \ell_1(t)=P+Dt~\&~\ell_2(s)=Q+Es\) are two skew lines. That means not \(\displaystyle D \| E\) and \(\displaystyle \ell_1\cap\ell_2=\emptyset\)

Now there is one and only one line the is perpendicular to \(\displaystyle \ell_1~\&~\ell_2\)

Use this reference to find out how to write that line.