Answer:IDK
Step-by-step explanation:I have the SAME QUESTION.
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
Answer:
The statement is false
Step-by-step explanation:
we know that
The tangent function will be positive when the sine function and the cosine function have the same sign
so
In the first quadrant the tangent function is positive
In the third quadrant the tangent function is positive
so
The statement is false
Answer:
y=-2x+2
Step-by-step explanation:
the slope intercept form is y=mx+b
slope - to find the slope from 2 point you do (y2-y1)/(x2-x1) so it would be
(14-2)/(-6-0). which out simplify to -2.
y-intercept - the y-intercept is given in one of the points (0,2).
