Addition of vectors 2

Algebra Level 1

For the three vectors a = ( 2 , 2 , 4 ) b = ( 0 , 8 , 4 ) c = ( 20 , 4 , 8 ) , \begin{aligned} a &= (2, -2, 4) \\ b &= (0, 8, 4) \\ c &= (-20, -4, 8), \end{aligned} what is the value of 4 a + 12 b 3 c ? 4a + 12b - 3c?

( 68 , 100 , 40 ) (68, 100, 40) ( 38 , 43 , 13 ) (38, 43, 13) ( 54 , 30 , 20 ) (54, 30, -20) ( 43 , 65 , 84 ) (43, 65, 84)

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2 solutions

Saurabh Mallik
May 12, 2014

We have to take one value for each variable at a time.

a = ( 2 , 2 , 4 ) , b = ( 0 , 8 , 4 ) , c = ( 20 , 4 , 8 ) a=(2,-2,4), b=(0,8,4), c=(-20,-4,8)

So, the first value of 4 a + 12 b 3 c 4a+12b-3c is:

4 a + 12 b 3 c = 4 × 2 + 12 × 0 3 × ( 20 ) 4a+12b-3c=4\times2+12\times0-3\times(-20)

4 a + 12 b 3 c = 8 + 0 + 60 4a+12b-3c=8+0+60

4 a + 12 b 3 c = 68 4a+12b-3c=68

Second value of 4 a + 12 b 3 c 4a+12b-3c is:

4 a + 12 b 3 c = 4 × ( 2 ) + 12 × 8 3 × ( 4 ) 4a+12b-3c=4\times(-2)+12\times8-3\times(-4)

4 a + 12 b 3 c = 8 + 96 + 12 4a+12b-3c=-8+96+12

4 a + 12 b 3 c = 100 4a+12b-3c=-100

Third value of 4 a + 12 b 3 c 4a+12b-3c is:

4 a + 12 b 3 c = 4 × 4 + 12 × 4 3 × 8 4a+12b-3c=4\times4+12\times4-3\times8

4 a + 12 b 3 c = 16 + 48 24 4a+12b-3c=16+48-24

4 a + 12 b 3 c = 40 4a+12b-3c=40

Thus, the answer is: 4 a + 12 b 3 c = ( 68 , 100 , 40 ) 4a+12b-3c=\boxed{(68,100,40)}

Vikas Sharma
May 26, 2014

very easy pbm............

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