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Question
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"r" = "x"bar"a" + "y"bar"b" + "z"bar"c"` ...(i) where x, y, z are scalars
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)]
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