The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k
Let r be the vector perpendicular to A and B,
r = A * B
A = 3i + 6j - 2k
B = 4i - j + 3k
a1 = 3
a2 = 6
a3 = - 2
b1 = 4
b2 = - 1
b3 = 3
a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k
a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k
a * b = 16 i - 17 j - 27 k
The perpendicular vector, r = 16 i - 17 j - 27 k
Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k
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Answer:
Given:
m=1000kg
u= 16.7m/s
v=0m/s
F=8000N
Required:
s=?
Solution:
F=m × a
8000N=1000kg × a
a=8m/s^2
Since it decelerate a= -8m/s^2
v^2 = u^2 + 2as
s=v^2 - u^2 / 2a
s= 0 - (16.7m/s)^2 / 2 × -8m/s^2
s= -278.89/-16
s= 17.43m
The car travels approximately 17.43m before it stops
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