00:01
In part a, we have as to find the parametric equations of the line that intersects the plane x plus y plus z is equal to 2 and the plane x plus 5y plus 5 z is equal to 2.
00:20
Let's consider these two equations as equation number 1 and equation number 2.
00:26
Understand that the required parametric equation of the line is given by the formula.
00:32
Of t is equal to x1 plus a t, y of t is equal to y1 plus b t and z of t is equal to z 1 plus c t.
00:45
In our case, x1 comma y 1 comma z 1 is it 1 is nothing but the point that is on the intersecting line.
00:58
Similarly, the point a comma b comma c is nothing but direction.
01:04
Vector of the intersecting line.
01:13
So first let's compute the value of a comma b comma c.
01:17
Let's consider the normal vectors of the two planes.
01:25
So n1 will be equal to the coefficient of x, y is at equation number 1.
01:32
So it is nothing but 1 comma 1 comma 1 .1.
01:36
Similarly n2 is equal to 1 comma 5 comma 5 these are the normal vectors of the given plane we need to compute the cross product of these two vectors so it will be equal to determinant of the matrix whose elements are i j k 1 1 1 the elements in n 1 and the third row will be 1 5 5 if we compute the determinant of this matrix we will get i times pi minus pi minus y minus j times five minus one plus k times five minus five minus one all these are vectors so it will be equal to zero i minus four j plus four king so we can write this as zero comma minus four comma four so this is the value of a comma v comma c now we need to find the value of x1y 1 z1 so we need to solve the equations 1 and 2 to find the intersecting points if we subtract equation number 2 from equation number 1 we will get minus 4y minus 4 z is equal to 0 which implies that the value of y will be equal to minus z and from equation number 1 if we substitute this expression in equation number 1 we will get x minus z plus z is equal to 2 which gives the value of x as 2.
03:14
So any point on the intersecting line will look like 2 comma y comma minus y.
03:24
So let's assume x1 comma y 1 comma z 1 is equal to 2 .1 1 minus 1 in our case.
03:34
So let's substitute the value of x1 y 1 z 1 and a comma b comma c in the parametric equation formula.
03:44
A comma b comma c is nothing but 0 minus 4 comma 4.
03:48
So x of t will be equal to 2 plus 0 t which is equal to 2.
03:54
Y of t will be equal to 1 minus 4 t and z of t will be equal to minus 1 plus 4t.
04:05
So this is the answer for the first part of the problem...