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  • If a,b,c determine the vertices of a triangle, show that 1/2 b x c +c gives the vector area of the triangle. Hence deduce the condition that the three points a, b, c are collinear. Also find the unit vector normal to the plane of the triangle.

If \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}  determine the vertices of a triangle, show that \frac{1}{2}[\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}]  gives the vector area of the triangle. Hence deduce the condition that the three points a, b, c are collinear. Also find the unit vector normal to the plane of the triangle.

Answers (1)

Let \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}  are vertices of a triangle ABC.

Also, we get

Position vector of A=\overrightarrow{\mathrm{a}}

Position vector of B=\overrightarrow{\mathrm{b}}

Position vector of C=\overrightarrow{\mathrm{c}}

We need to show that,

\frac{1}{2}[\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}]  gives the vector are of the triangle.

We know that,

Vector area of triangle ABC is given as,
\\ Area of \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$\\ Here, $\overrightarrow{\mathrm{AB}}=$ Position vector of $\mathrm{B}-$ Position vector of $\mathrm{A}$\\ $\Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$\\ $\overrightarrow{\mathrm{AC}}=$ Position vector of $\mathrm{C}-$ Position vector of $\mathrm{A}$\\

\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}} \\ \therefore \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \times(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}})| \\ \Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}| \\ {[\because-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}},-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0]}

\Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\ldots(\mathrm{j})}

Thus, shown.

We know that, two vectors are collinear if they lie on the same line or parallel lines.

For \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}  to be collinear, area of the ?ABC should be equal to 0.

⇒ Area of ?ABC = 0

\\ \Rightarrow \frac{1}{2}|\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}|=0 \\ \Rightarrow \vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}=0

Thus, this is the required condition for \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}  to be collinear.

Now, we need to find the unit vector normal to the plane of the triangle.

Let \vec{\pi}  be the unit vector normal to the plane of the triangle.

\\ \overrightarrow{\mathrm{n}}=\frac{\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|}$ \\Note that, $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}_{\text {from equation }(\mathrm{i})}$ \\And, $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\text {from equation }(\mathrm{i})}$

So, \overrightarrow{\mathrm{n}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}$

Thus, unit vector normal to the plane of the triangle is \frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}$

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