If a , b , c are non-zero, non-coplanar vectors, prove that the v…

Question

If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-zero, non-coplanar vectors, prove that the vector is coplanar:
$5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}},\ 7\vec{\text{a}}-8\vec{\text{b}}+9\vec{\text{c}}$ and $3\vec{\text{a}}+20\vec{\text{b}}+5\vec{\text{c}}$

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Answer

We know that, Three vectors are coplanar if one of the vector can be expressed as the linear combination of other two. Let, $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}}\\=\text{x}\big(7\vec{\text{a}}-8\vec{\text{b}}+9\vec{\text{c}}\big)+\text{y}\big(3\vec{\text{a}}+20\vec{\text{b}}+5\vec{\text{c}}\big)$ $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}}\\=7\vec{\text{a}}\text{x}-8\vec{\text{b}}\text{x}+9\vec{\text{c}}\text{x}+3\vec{\text{a}}\text{y}+20\vec{\text{b}}\text{y}+5\vec{\text{c}}\text{y}$ $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}}\\=\big(7\text{x}+3\text{y}\big)\vec{\text{a}}+\big(-8\text{x}+20\text{y}\big)\vec{\text{b}}+\big(9\text{x}+5\text{y}\big)\vec{\text{c}}$ Comparing the LHS and RHS, 7x + 3y = 5 .....(i) -8x + 20y = 6 .....(ii) 9x + 5y = 7 .....(iii) For solving (i) and (ii), Subtract -8 × (i) from 7 × (ii),

$\text{y}=\frac{82}{164}$ $\text{y}=\frac{1}2$ Put $\text{y}=\frac{1}2$ in equation (i), $7\text{x}+3\text{y}=5$ $7\text{x}+3\Big(\frac{1}2\Big)=5$ $7\text{x}+\frac{3}2=5$ $7\text{x}=\frac{5}1-\frac{3}2$ $7\text{x}=\frac{10-3}2$ $7\text{x}=\frac{7}2$ $\text{x}=\frac{7}{14}$ $\text{x}=\frac{1}{2}$ Now, put $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}2$ in equation (iii), $9\text{x}+5\text{y}=7$ $9\Big(\frac{1}2\Big)+5\Big(\frac{1}2\Big)=7$ $\frac{9}2+\frac{5}2=7$ $\frac{14}2=7$ $7=7$LHS = RHS
$\therefore$ The value of x, y satisfy equation (iii).
So, $5\vec{\text{a}}+6\vec{\text{b}}+7\vec{\text{c}},\ 7\vec{\text{a}}-8\vec{\text{b}}+9\vec{\text{c}},\ 3\vec{\text{a}}+20\vec{\text{b}}+5\vec{\text{c}}$ are coplanar.