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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors4 Marks
Question
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-zero, non-coplanar vectors, prove that the vector is coplanar:
$\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ -3\vec{\text{b}}+5\vec{\text{c}}$ and $-2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$
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Answer

We know that,

Three vectors are coplanar if one of them can be expressed as the linear combination of other two.

Let,

$\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}}\\=\text{x}\big(-3\vec{\text{b}}+5\vec{\text{c}}\big)+\text{y}\big(-2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}\big)$

$\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}}\\=-3\vec{\text{b}}\text{x}+5\vec{\text{c}}\text{x}+2\vec{\text{a}}\text{y}+3\vec{\text{b}}\text{y}-4\vec{\text{c}}\text{y}$

$\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}}\\=\big(-2\text{y}\big)\vec{\text{a}}+\big(-3\text{x}+3\text{y}\big)\vec{\text{b}}+\big(5\text{x}-4\text{y}\big)\vec{\text{c}}$

Comparing the LHS and RHS,

-2y = 1 .....(i)

-3x + 3y = -2 .....(ii)

5x - 4y = 3 .....(iii)

For solving (i) and $\text{y}=-\frac{1}2$

Put value of y in equation (ii),

$-3\text{x}+3\text{y}=-2$

$-3\text{x}+3\Big(-\frac{1}2\Big)=-2$

$-3\text{x}-\frac{3}{2}=-2$

$-3\text{x}=\frac{-2}1+\frac{3}2$

$-3\text{x}=\frac{-4+3}2$

$-3\text{x}=\frac{-1}2$

$\text{x}=\frac{-1}{-6}$

$\text{x}=\frac{1}6$

Now, put the value of x and y in equation (iii),

$5\text{x}-4\text{y}=3$

$5\Big(\frac{1}6\Big)-4\Big(-\frac{1}2\Big)=3$

$\frac{5}6+\frac{4}2=3$

$\frac{5+12}6=3$

$\frac{17}6=3$

$\text{LHS}\neq\text{RHS}$

So, value of x and y do not satisfy the equation (iii).

So,

vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ -3\vec{\text{b}}+5\vec{\text{c}},\ -2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ are not coplanar.

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