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:
23.5(to the nearest tenth)
Step-by-step explanation:
pythagoras theorem states
c^2= a^2+ b^2
a= 23
b=5
c^2= 529+25
c^2= 554
c= square root of 554
c = 23.5 (approximately)
this took me a lot of time
giving me brainliest will make me happy
Answer:
The Hypotenuse is side just opposite to the right angle.
Step-by-step explanation:
By Pythagoras' theorem,
c^2=a^2+b^2
c^2(hypotenuse to be found)= 30^2+ 16^2
<em>c^2=1156
</em>
<em><u>Therefore, c ( i.e. the hypotenuse) will be the square root of 1156. </u></em>
<em><u>The square root of 1156 is 34. </u></em>
<h2><em><u>
Thus, c ( hypotenuse) = 34</u></em></h2>
Hope it helped :)
Answer:
ummmm 90% sure its 13/10.......?
Step-by-step explanation:
:/
1.4 is the answer pretty sure
