Video Transcript
Determine to the nearest second the
measure of the angle between the two lines that have direction ratios of negative
four, negative three, negative four and negative three, negative three, negative
one.
We’re given the direction ratios of
two lines. Let’s call our lines 𝐿 one and 𝐿
two. And we’re asked to find the measure
of the angle between these two lines. Assuming for the moment that our
direction vectors are in the positive sense to each other, then the angle between
them is an acute angle. Let’s call that 𝜃. Now, recall that the direction
ratios of a line, and we’re given two here, are the 𝑥-, 𝑦-, and 𝑧-coefficients of
the unit vectors 𝐢, 𝐣, and 𝐤 of the direction vector of the line so that the
direction vector is defined as 𝐝 is equal to 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤 where 𝑥, 𝑦,
and 𝑧 are the direction ratios.
In our case then, our direction
vectors are 𝐝 one is negative four 𝐢 plus negative three 𝐣 plus negative four 𝐤
and 𝐝 two is negative three 𝐢 plus negative three 𝐣 plus negative one times
𝐤. Now, to find the angle between two
lines with direction vectors 𝐝 one and 𝐝 two, respectively, we can use the formula
cos 𝜃 is the scalar product of the two direction vectors 𝐝 one and 𝐝 two divided
by the product of the magnitudes of the two direction vectors. And remember that the scalar
product of two vectors is the sum of the like-for-like products of the coefficients
of the unit vectors 𝐢, 𝐣, and 𝐤 of each vector, and this is a scalar. And remember that the magnitude of
a vector is the square root of the sum of the squares of the coefficients of 𝐢, 𝐣,
and 𝐤, the unit vectors.
In the case of a direction vector
of course, the coefficients of 𝐢, 𝐣, and 𝐤 are the direction ratios. So in our case, our scalar product
is negative four times negative three plus negative three times negative three plus
negative four times negative one. That is 12 plus nine plus four,
which is 25. To use our formula, we also need to
find our magnitudes. Now the magnitude of 𝐝 one, the
first direction vector, is the square root of negative four squared plus negative
three squared plus negative four squared. That is the square root of 16 plus
nine plus 16, which is the square root of 41. And the magnitude of 𝐝 two is the
square root of negative three squared plus negative three squared plus negative one
squared. That’s the square root of nine plus
nine plus one, which is the square root of 19.
So we have the scalar product of
our two direction vectors is 25. And the magnitude of 𝐝 one is the
square root of 41. The magnitude of 𝐝 two is the
square root of 19. And making some room, we can put
these into our formula so that cos 𝜃 is 25 over the square root of 41 times the
square root of 19. Taking the inverse cos on both
sides, we have 𝜃 is the inverse cos of 25 over the square root of 41 times the
square root of 19. And this gives us an angle of 𝜃 is
26.399 and so on degrees. But we’re not quite finished yet
since we’re asked for the measure of the angle between our two lines to the nearest
second.
At this point, it’s important we
haven’t rounded after the decimal point. We have 26 degrees, and next we
find our minutes by multiplying everything after the decimal point by 60. And this gives us 23.952 etcetera
minutes. Again, we don’t round after the
decimal point because in order to find the seconds, we multiply everything after the
decimal point again by 60 which gives us 57.126. So we have 26 degrees, 23 minutes,
and 57 seconds. And our angle is acute, so we have
the right directions. And so the measure of the angle
between the two lines that have direction ratios of negative four, negative three,
negative four and negative three, negative three, negative one to the nearest second
is 𝜃 is 26 degrees, 23 minutes, and 57 seconds.
