WEBVTT
00:00:01.000 --> 00:00:10.720
Given that π, π, and π are the direction cosines of a straight line, find the value of π squared plus π squared plus π squared.
00:00:11.760 --> 00:00:15.200
Letβs just consider that we have the space π₯, π¦, and π§.
00:00:15.880 --> 00:00:19.390
Point π΄ is found at lowercase π₯π¦π§.
00:00:19.760 --> 00:00:23.930
And weβll have a line from the origin to point π΄, like this.
00:00:24.780 --> 00:00:35.440
Line ππ΄ is made up of three different vectors, here π₯ in pink, π¦ in yellow, and π§ in green.
00:00:36.040 --> 00:00:48.710
Under these conditions, we can let π be equal to the cos of πΌ, where πΌ is the angle measure between this pink line and our vector ππ΄.
00:00:49.560 --> 00:00:57.170
π is gonna be equal to cos of π½, where π½ is the angle measure between the yellow line and blue line here.
00:00:57.770 --> 00:01:05.400
π [π] will be equal to the cosine of πΎ, where πΎ is the angle between the green line and the blue line in our figure.
00:01:06.040 --> 00:01:20.770
And the cosine value will be equal to the adjacent side length π₯ over the length of our line, which we find by saying π₯ squared plus π¦ squared plus π§ squared and then taking the square root of that value.
00:01:21.710 --> 00:01:25.550
To find the cos of B [π½], weβll follow the same procedure.
00:01:25.770 --> 00:01:28.720
But this time, weβll have the length π¦ as the numerator.
00:01:29.540 --> 00:01:36.490
And the cos of πΎ will be equal to π§ over the square root of π₯ squared plus π¦ squared plus π§ squared.
00:01:37.250 --> 00:01:41.880
Remember, weβre looking for π squared plus π squared plus π squared.
00:01:42.460 --> 00:01:49.720
If we plug in the values that we know for π, π, and π, we need to take π₯ squared.
00:01:50.160 --> 00:01:55.800
And then, we need to square the square root of π₯ squared plus π¦ squared plus π§ squared.
00:01:56.590 --> 00:02:02.140
And that will give us π₯ squared over π₯ squared plus π¦ squared plus π§ squared.
00:02:02.920 --> 00:02:08.460
When we square π, we get π¦ squared over π₯ squared plus π¦ squared plus π§ squared.
00:02:08.950 --> 00:02:14.110
And when we square π, we get π§ squared over π₯ squared plus π¦ squared plus π§ squared.
00:02:15.100 --> 00:02:20.050
All three of these fractions have a common denominator, which means we can add their numerators.
00:02:20.590 --> 00:02:24.570
π₯ squared plus π¦ squared plus π§ squared canβt be simplified.
00:02:24.570 --> 00:02:28.280
So we just leave it π₯ squared plus π¦ squared plus π§ squared.
00:02:28.960 --> 00:02:30.790
And the denominator doesnβt change.
00:02:30.820 --> 00:02:34.280
Itβs also π₯ squared plus π¦ squared plus π§ squared.
00:02:35.000 --> 00:02:40.620
And when the numerator equals the denominator, the value of that fraction is one.
00:02:41.460 --> 00:02:45.750
Weβve just shown that π squared plus π squared plus π squared equals one.
00:02:46.230 --> 00:02:53.260
In fact, the squares of the direction cosines of a straight line when added together always equal one.