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The position vector of the point $A$ is $2\vec{i} - \vec{k}$ and the equation of the the line is:

$$\vec{r} = (-7, 15, -5) + \lambda (3, -7, 4) \ . $$

Find the position vectors of points $B$ and $C$, both lying on the line, such that the length $\overline{AB} = \overline{AC} = 10$.

So this is what I've done so far:

I found the foot of the perpendicular from point $A$ to the line and labeled that $M$. Using the fact that point $M$ lies on the line and that it is perpendicular to the director vector of the line, I found the coordinate of point $M$ (-1, 1, 3). Now that I have the position vector of point $M$ and the magnitude of $\overline{AC}$, I can find the magnitude of $\overline{MC}$ using the Pythagoras Theorem but I don't know how to continue from there and or if I am at all on the right path.

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  • $\begingroup$ Sorry about that but I don't know how to edit the question so that matches the quality standards. I was simply having some trouble with some homework and I thought I would try posting here as I couldn't find a similar question being answered anywhere else $\endgroup$ – Aatrix Apr 14 at 11:19
  • $\begingroup$ Please say what have you tried. For instance, can you write the formula for the distance between $A$ and a point on the line for $\lambda$ unknown? $\endgroup$ – Ertxiem Apr 14 at 11:39
  • $\begingroup$ Did you look at the link in the first comment that said it would "help you recognize and resolve the issues"? It leads to more links; did you look at this one? By showing some effort, even if it goes nowhere or leads you in a circle back to where you started, you give us clues about what kind of help to give. $\endgroup$ – David K Apr 14 at 11:44
  • $\begingroup$ Hopefully it is better now. Sorry about messing up this is my first time. And thank you @Ertxiem for suggesting the edit $\endgroup$ – Aatrix Apr 14 at 12:09
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You can solve it the way you are working, but it seems to me that it's going to give you a bit more work that just solving: $$ \overline{AR}^2 = 10^2$$ with $R=( -7 + 3 \lambda, 15 -7 \lambda, -5 + 4 \lambda )$.

It will give a quadratic polynomial for $\lambda$, hence the two solutions.

Can you solve this?


Edit:

Since $A = (2, 0, -1)$, then the square of the distance between $A$ and the generic point of the straight line $r$ is:

$$ \overline{AR}^2 = (-7 + 3 \lambda - 2)^2 + (15 -7 \lambda - 0)^2 + (-5 + 4 \lambda - 1)^2 $$

Since we want $\overline{AR}^2 = 10^2$, we get $$ (-7 + 3 \lambda - 2)^2 + (15 -7 \lambda - 0)^2 + (-5 + 4 \lambda - 1)^2 = 100 $$

Which gives: $$ (3 \lambda - 9)^2 + (15 -7 \lambda)^2 + (4 \lambda - 6)^2 = 100 $$ $$ 9 \lambda^2 - 54 \lambda + 81 + 225 -210 \lambda + 49 \lambda^2 + 16 \lambda^2 -24 \lambda + 36 = 100 $$

Now, you just need to solve the quadratic polynomial in $\lambda$.

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  • $\begingroup$ Nope not really. I don't understand what $R$ represents, why you squared the expression and how you would get a quadratic polynomial from that. I should be able to solve for $\lambda$ and find the position vectors of $B$ and $C$ from there though $\endgroup$ – Aatrix Apr 14 at 12:46
  • $\begingroup$ So by subbing $R$ into $\overline{AR}$ i got $( 81 - 54 \lambda +9 \lambda^2, 225 - 210 \lambda +49 \lambda^2, 36 - 48 \lambda +16 \lambda^2 ) $ = 100. How do I continue? Do i just add all of them up and set them equal to 100 and solve for $\lambda$? $\endgroup$ – Aatrix Apr 14 at 13:03
  • $\begingroup$ The distance is the sum of the square of difference for each coordinate. I added those steps in my answer. $\endgroup$ – Ertxiem Apr 14 at 13:57
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You’re most of the way there. Once you’ve used the Pythagorean theorem to determine the distance of the two points from $M$, you just need to move that far along the line in both directions. The direction vector given in the parametric equation of the line is $(3,-7,4)$, so you just have to find a suitable multiple of this that has the required length.

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