Find the position vectors of points B and C, both lying on a line, such that the length AB = AC = 10

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.

• 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 – Aatrix Apr 14 at 11:19
• 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? – Ertxiem Apr 14 at 11:39
• 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. – David K Apr 14 at 11:44
• Hopefully it is better now. Sorry about messing up this is my first time. And thank you @Ertxiem for suggesting the edit – Aatrix Apr 14 at 12:09

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 $$\lambda$$.

• 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 – Aatrix Apr 14 at 12:46
• 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 $\lambda$? – Aatrix Apr 14 at 13:03
• The distance is the sum of the square of difference for each coordinate. I added those steps in my answer. – Ertxiem Apr 14 at 13:57

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.