Find a.b and a x b if a=i+2j+k and |b|=3 acting along c=i+j+k.?

Answer 1

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a= i + 2j + k

|b| = magnitude of b = 3

Firstly; the magnitude of a vector α, is given by:

|α| = √[(α1)² + (α2)²],

so for example, the magnitude of |a| is:

|a| = √[(1)² + (2)² + (1)²] = √[1 + 4 +1] = √6 = 2.45

also the dot product is given by:

a.b = a1b1 + a2b2 + a3b3

now onto the answer:

vector c has a direction of (i+j+k)/√3

and since vector b is along c vector b = 3(i+j+k)/√3 = √3(i+j+k)

a.b = [i + 2j + k] . [ √3(i+j+k) ]

applying the above formula:

(1 x [√3 x 1]) + (2 x [√3 x 1]) + (1 x [√3 x 1]) = √3 + [2 x √3 ] + √3 = 6.928 or = 4√3

now the cross product utilises a matrix (which is hard to do here, but im will provide the formula); it states:

axb = <a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1>

axb =

| i j k |

| 1 2 1 |

| √3 √3 √3 |

=

i | (2 x √3 ) - (1 x √3 ) | - j | (1 x √3) - (√3 x 1)| + k | (1 x √3) - ( 2 x √3) |

= √3 i - 0j + (-)√3 k

= √3 i - √3 k

= √3 (i-k)

summary:

a.b = 4√3

axb = √3 (i-k)