# Express -i^-3j^+4k^ as the linear combination of the vectors 2i^+j^-4k^,2i^-j^+3k^ and 3i^+j^-2k^ - Mathematics and Statistics

Sum

Express - hat"i" - 3hat"j" + 4hat"k" as the linear combination of the vectors 2hat"i" + hat"j" - 4hat"k", 2hat"i" - hat"j" + 3hat"k" and 3hat"i" + hat"j" - 2hat"k"

#### Solution

Let bar"a" = 2hat"i" + hat"j" - 4hat"k",

bar"b" = 2hat"i" - hat"j" + 3hat"k",

bar"c" = 3hat"i" + hat"j" - 2hat"k"

bar"r" = - hat"i" - 3hat"j" + 4hat"k"

Suppose bar"p" = "x"bar"a" + "y"bar"b" + "z"bar"c".

Then, - hat"i" - 3hat"j" + 4hat"k" = "x"(2hat"i" + hat"j" - 4hat"k") + "y"(2hat"i" - hat"j" + 3hat"k") + "z"(3hat"i" + hat"j" - 2hat"k")

∴ - hat"i" - 3hat"j" + 4hat"k" = (2"x" + 2"y" + 3"z")hat"i" + ("x" - "y" + "z")hat"j" + (- "4x" + "3y" - "2z")hat"k"

By equality of vectors, we get

2x + 2y + 3z = −1

x − y + z = −3

−4x + 3y − 2z = 4

We have to solve these equations by using Cramer’s Rule.

D = |(2,2,3),(1,-1,1),(-4,3,-2)|

= 2(2 − 3) − 2(− 2 + 4) + 3(3 − 4)

= 2(–1) – 2(2) + 3(–1)

= −2 − 4 − 3

= −9 ≠ 0

Dx = |(-1,2,3),(-3,-1,1),(4,3,-2)|

= −1(2 − 3) − 2(6 − 4) + 3(− 9 + 4)

= – 1(– 1) – 2(2) + 3(– 5)

= 1 − 4 − 15

= −18

Dy = |(2,-1,3),(1,-3,1),(-4,4,-2)|

= 2(6 − 4) + 1(− 2 + 4) + 3(4 − 12)

= 2(2) + 1(2) + 3(– 8)

= 4 + 2 − 24

= −18

Dz = |(2,2,-1),(1,-1,-3),(-4,3,4)|

= 2(− 4 + 9) − 2(4 − 12) − 1(3 − 4)

= 2(5) – 2(– 8) – 1(–1)

= 10 + 16 + 1

= 27

∴ x = "D"_"x"/"D" = (- 18)/-9 = 2

∴ y = "D"_"y"/"D" = (- 18)/-9 = 2

∴ z = "D"_"z"/"D" = 27/-9 = - 3

∴ bar"r" = 2bar"a" + 2bar"b" - 3bar"c"  .......[From (i)]

Concept: Vectors and Their Types
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