Video Transcript
Find the equation of the line through the origin that intersects the line 𝐫 sub one is equal to the vector negative one, two, three plus 𝑡 sub one multiplied by the vector three, negative five, one orthogonally.
In this question, we’re asked to find the equation of the line and we’re told some information about this line. First, we’re told that this line passes through the origin, and second we’re told it intersects a given line 𝐫 sub one orthogonally. Since we’re told that our line passes through the origin, we know it passes through the point with coordinates zero, zero, zero. And so if we wanted to write the equation of this line in vector form, we could use the position vector zero, zero, zero as our position vector 𝐫 sub zero.
We know that the equation of this line is of the form 𝐫 sub two is equal to the vector zero, zero, zero plus 𝑡 sub two multiplied by the direction vector 𝐝 sub two. And of course, adding the zero vector won’t change this equation. So we can rewrite this as 𝐫 sub two is equal to 𝑡 sub two multiplied by 𝐝 sub two. In other words, we only need to find the direction vector of this line to determine its equation.
And to find the direction vector of this line, we’re going to use the fact that it intersects the given line orthogonally. And we can recall this is the same as saying that they intersect at right angles. And we can also recall if they have direction vectors 𝐝 sub one and 𝐝 sub two, then the dot product of these two vectors must be zero. And we’re given the direction vector of the first line in the question; it’s the vector three, negative five, one. So we’ve shown the dot product between the vector three, negative five, one and 𝐝 sub two is equal to zero. But this is not the only equation we can find involving the direction vector 𝐝 sub two.
Remember, we are also told in the question that our line intersects the given line. And to help us see how this might be useful, let’s sketch a diagram. We want to consider a point 𝑃 on the line 𝐿 sub one, where 𝐿 sub one is the line given to us in the question. Let’s say the line we’re looking for intersects 𝐿 sub one at the point 𝑃. We also know it passes through the origin. We’re told this in the question. We now have two distinct points on our line 𝐿 sub two. We can use this to find an expression for its direction vector. This gives us one possible direction vector of our line, the vector from 𝑂 to 𝑃.
However, we don’t yet know the coordinates of 𝑃 because we don’t know the point of intersection between the two lines. All we know is that 𝑃 lies on the line 𝐿 sub one, so its position vector satisfies the vector equation of the line. And since the position vector of 𝑃 is the same as the vector from 𝑂 to 𝑃, we get the vector from 𝑂 to 𝑃 is equal to the vector negative one, two, three plus 𝑡 sub one multiplied by the vector three, negative five, one since 𝑃 lies on the line 𝐿 sub one. We can now distribute our scalar 𝑡 sub one over the vector and then add the two vectors together. We get the vector from 𝑂 to 𝑃 is equal to the vector negative one plus three 𝑡 sub one, two minus five 𝑡 sub one, three plus 𝑡 sub one.
But remember, we’ve already said the vector from 𝑂 to 𝑃 is a direction vector of our line. And in particular because the lines are orthogonal, the dot product between their direction vectors must be zero. Therefore, the dot product between the vector three, negative five, one and the vector from 𝑂 to 𝑃 must be zero. Therefore, we get the dot product between the vectors three, negative five, one and negative one plus three 𝑡 sub one, two minus five 𝑡 sub one, three plus 𝑡 sub one is equal to zero.
And now, we can find an equation for 𝑡 sub one by evaluating this dot product. We do this by recalling the dot product between two vectors of equal dimensions is the sum of the products of the corresponding components. So evaluating the dot product, we get three multiplied by negative one plus three 𝑡 sub one plus negative five times two minus five times 𝑡 sub one plus one multiplied by three plus 𝑡 sub one. And now, all that’s left to do is simplify this expression. We start by noting multiplying by one doesn’t change the value, and we’ll also distribute over the other two parentheses. This then gives us that zero is equal to negative three plus 9𝑡 sub one minus 10 plus 25𝑡 sub one plus three plus 𝑡 sub one.
We can then simplify and rearrange this equation. Negative three plus three is equal to zero. Then we add 10 to both sides of the equation and collect like terms. We get 10 is equal to 35𝑡 sub one. And finally, we divide both sides of the equation through by 35 and simplify. We get 𝑡 sub one is two-sevenths. We can now substitute 𝑡 sub one is equal to two over seven into the vector equation of 𝐿 sub one to find the position vector of 𝑃. Or alternatively, we can just substitute it into our expression for the vector from 𝑂 to 𝑃. Substituting 𝑡 sub one is equal to two over seven into our equation, we get the vector from 𝑂 to 𝑃 is the vector negative one plus three times two over seven, two minus five times two over seven, three plus two over seven.
And we can now evaluate each component separately. We get the vector from 𝑂 to 𝑃 is the vector negative one over seven, four over seven, 23 over seven. And this is enough to answer our question. Remember, this is a possible direction vector of the line. However, we can make this easier. Remember, we can multiply direction vectors of lines by any nonzero scalar, and we can see every component of this vector has a denominator of seven.
So let’s instead choose our direction vector of the line to be the vector 𝐎𝐏 multiplied by seven. And to multiply this vector by this scalar, we multiply all of the components by seven. We get the vector negative one, four, 23, and we’ll use this as our direction vector of our line 𝐫 sub two, which then gives us our final answer. The vector equation of the line through the origin that intersects the line 𝐫 sub one is equal to the vector negative one, two, three plus 𝑡 sub one times the vector three, negative five, one orthogonally is vector 𝐫 sub two is equal to 𝑡 sub two multiplied by the vector negative one, four, 23.
