Video Transcript
If the straight line π₯ plus eight
over negative 10 is equal to π¦ plus eight over π which is equal to π§ plus 10 over
negative eight is perpendicular to π₯ plus five over negative four which is equal to
π¦ plus eight over 10 and π§ equals eight, find π.
We notice that both our straight
lines in this question are given in symmetric form. π₯ minus π₯ sub zero over π is
equal to π¦ minus π¦ sub zero over π which is equal to π§ minus π§ sub zero over
π. Any straight line in this form can
be rewritten in vector form π₯ sub zero, π¦ sub zero, π§ sub zero plus π‘ multiplied
by π, π, π. π₯ sub zero, π¦ sub zero, π§ sub
zero is the initial position and π, π, π is the direction of the straight
line.
The vector equation of our first
line is therefore equal to negative eight, negative eight, negative 10 plus π‘
multiplied by negative 10, π, negative eight. The second equation doesnβt appear
to have a π§-component; however, we are told that π§ equals eight. Therefore, the initial position is
negative five, negative eight, eight. As the π§-component is remaining
constant, the direction of the straight line is negative four, 10, zero. We know that when two straight
lines are perpendicular, the scalar product of the directions equals zero. To calculate the scalar products,
we find the sum of the products of the individual components.
This gives us negative 10
multiplied by negative four plus π multiplied by 10 plus negative eight multiplied
by zero. This must sum to zero. Negative 10 multiplied by negative
four is equal to 40, π multiplied by 10 is 10π, and negative eight multiplied by
zero is zero. Subtracting 40 from both sides of
this equation gives us 10π is equal to negative 40. We can then divide both sides of
this equation by 10, giving us π is equal to negative four. If the two straight lines are
perpendicular, the value of π is equal to negative four.
