if i,j,k are the standard unit basic vectors, in 3 space, determine the value of
k dot(j-3k)+(i-4k) dot (i-4k)-8 |i cross -k|
(Simplify without using components)
Writing in component notation,
<1,0,0>=i
<0,1,0>=j
<0,0,1>=k
and using rules
ixj=k, jxk=i, kxi=j
jxi=-k, kxj=-i, ixk=-j
and the magnitude of all unit vectors is always one.
Above reduces to
<0,0,1>.<0,1,-3> + <1,0,-4>.<1,0,-4> - 8|ixk|
you can complete the computations and post your answer for a check if you wish.
thanks! This is my answer, is it correct?
(0,0,1).((0,1,-3) +(1,0,-4).(1,0,-4)-8|(1,0,0)x(0,0,-1)|
=(0)(0)+(0)(1)+(1)(-3)+(1)(1)+(0)(0)+(-4)(-4)-8|(0,1,0)|
=-3+1+16-8
=6
agree
was I supposed to multiple -8 with |ixk| as I did above or subtract eight from the dot products?
oh nevermind I get the same answer either way, thanks everyone
To simplify the expression without using components, we can use the properties of dot product and cross product.
Let's break down the expression step by step:
1. k dot (j-3k): The dot product of the standard unit vector k and the vector (j-3k) is equal to the product of their magnitudes multiplied by the cosine of the angle between them. Since k is a unit vector, its magnitude is 1. The vector (j-3k) has a magnitude of √(1^2+(-3)^2) = √10. The angle between k and (j-3k) is 90 degrees (since they are perpendicular). Therefore, k dot (j-3k) = 1 * √10 * cos(90) = √10 * 0 = 0.
2. (i-4k) dot (i-4k): The dot product of the vector (i-4k) with itself is equal to the square of its magnitude. The magnitude of (i-4k) is √(1^2+(-4)^2) = √17. Therefore, (i-4k) dot (i-4k) = (√17)^2 = 17.
3. |i cross -k|: The cross product of the vectors i and -k is another vector perpendicular to both i and -k. Its magnitude is equal to the product of the magnitudes of i and -k multiplied by the sine of the angle between them. Since i and -k are perpendicular, the angle between them is 90 degrees, and sin(90) = 1. The magnitudes of i and -k are both 1. Therefore, |i cross -k| = 1 * 1 * 1 = 1.
Now, substituting these values back into the original expression:
k dot (j-3k) + (i-4k) dot (i-4k) - 8 |i cross -k|
= 0 + 17 - 8 * 1
= 17 - 8
= 9
So, the value of the given expression is 9.