Video Transcript
Find to the nearest second the
measure of the angle between the two straight lines negative two 𝑥 is four 𝑦 is
equal to negative three 𝑧 and negative four 𝑥 is negative five 𝑦 is equal to two
𝑧.
We’re given two lines to find in
Cartesian form. And remember, the Cartesian form of
a line is 𝑥 minus 𝑥 one over 𝑎 is equal to 𝑦 minus 𝑦 one over 𝑏 is equal to 𝑧
minus 𝑧 one over 𝑐. And that’s where the point 𝐴 with
coordinates 𝑥 one, 𝑦 one, 𝑧 one lies on the line and 𝑎, 𝑏, and 𝑐 are the
direction ratios of the direction vector 𝐝, which is parallel to the line. Let’s begin by calling our lines 𝐿
one and 𝐿 two. And by comparing the three terms in
the general Cartesian form to our lines, we can determine the point on the line. But more importantly, we can find
the direction ratios 𝑎, 𝑏, and 𝑐 and hence the direction vector 𝐝 for each
line.
We can then use the formula shown
to find the cosine of the angle between the two lines and take the inverse cosine to
find the angle. So let’s look first at our line 𝐿
one. Comparing the expression in 𝑥 to
our Cartesian form, we have negative two 𝑥 is 𝑥 minus 𝑥 one over 𝑎. And separating our fraction on the
right-hand side, that gives us 𝑥 over 𝑎 minus 𝑥 one over 𝑎. And comparing coefficients, we have
negative two is equal to one over 𝑎 and zero is negative 𝑥 one over 𝑎. From our first equation, solving
for 𝑎, we have 𝑎 is negative one over two. So from our second equation, this
must mean that zero is 𝑥 one or 𝑥 one is equal to zero. And making some room, we have 𝑎 is
negative a half and 𝑥 one is equal to zero.
And doing the same thing for our
𝑦-term, we find that 𝑏 is one-quarter and 𝑦 one is equal to zero. And finally, for our 𝑧-terms, we
find that 𝑧 is negative a third and 𝑧 one is equal to zero. Now, remembering that 𝑥 one, 𝑦
one, and 𝑧 one are the coordinates of our point 𝐴, we have 𝐴 with coordinates
zero, zero, zero. And remembering our direction
vector is given by 𝐝 is 𝑎𝐢 plus 𝑏𝐣 plus 𝑐𝐤 so that our direction vector for
the line 𝐿 one is 𝐝 one, which is negative a half 𝐢 plus one over four 𝐣 plus
negative one-third 𝐤. And we’re going to call our point
𝐴 one to distinguish it from the point on the line 𝐿 two.
Now if we do the same thing for our
second line 𝐿 two, 𝐿 two passes through the point 𝐴 two with coordinates zero,
zero, zero and has direction vector 𝐝 two equal to negative one over four 𝐢 plus
negative one over five 𝐣 plus one over two 𝐤. All we need to use in vectors are
direction vectors in the formula for cos 𝜃. We’re going to need the scalar
product of our two direction vectors. And remember, that’s the sum of the
products of the coefficients of 𝐢, 𝐣, and 𝐤, the unit vectors, and that the
magnitude of the vector is the square root of the sum of the squares of the
coefficients of 𝐢, 𝐣, and 𝐤.
In our case then, the scalar
product is negative a half times negative a quarter plus one-quarter times negative
one-fifth plus negative one-third times one-half. That is one over eight plus
negative one twentieth plus negative one over six, which is negative 11 over
120. And making some room, we next need
to work out the magnitudes of our two vectors. The magnitude of 𝐝 one is the
square root of negative a half squared plus one-quarter squared plus negative a
third squared, which evaluates to the square root of 61 over 12. And the magnitude of 𝐝 two is the
square root of negative a quarter squared plus negative a fifth squared plus a half
squared. And this evaluates to the square
root of 141 divided by 20.
So now we can put our three values
into our formula. And recall that dividing by a
fraction is the same as multiplying by its reciprocal. We have negative 11 times 12 times
20 in the numerator and 120 times the square root of a 61 times the square root of
141 in the denominator. We can cancel a 12 in the numerator
and the denominator. And the result is 10 in the
denominator with the 20 in the numerator. And this gives us the cos of the
angle 𝜃 is negative 22 over the square root of 61 times the square root of 141. Now making some room, if we take
the inverse cos on both sides, we find in fact to three decimal places that 𝜃 is
103.722 degrees. But this is an obtuse angle and our
angle should be acute. So let’s take a look at our
direction vectors.
In this plot of our direction
vectors, we can see that the angle between them is obtuse. But when we refer to the angle
between two lines, we mean the angle sandwiched between the positive senses of both
vectors. But our vectors are in opposite
directions. Hence, the angle we found is the
larger of the two. And in fact, it’s the smaller acute
angle that we want. The angle we found corresponds to
the 𝛽 in this diagram. And the angle we want is 𝛼 which
is 180 minus 𝛽. In our case, that corresponds to
180 minus 103.722 and so on. That is 76.278 and so on
degrees.
We’re not quite finished yet,
however, since we want the measure of the angle to the nearest second. Multiplying successively after the
decimal place by 60, we have 76 degrees, 16 minutes, and 39 seconds. So to the nearest second, the angle
between the two lines 76 degrees, 16 minutes, and 39 seconds.
