# Examples On Vector Cross Product Set-2

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Example – 25

Let Ai, i = 1, 2, 3, 4 be the areas of the faces of a tetrahedron. Let $${\vec n_i}$$, i = 1, 2, 3, 4 be the outward drawn normals to the respective faces with magnitudes equal to the corresponding areas.

Prove that $${\vec n_1} + {\vec n_2} + {\vec n_3} + {\vec n_4} = \vec 0$$

Solution: There is no loss of generality in assuming one vertex A to be the origin $$\vec 0$$ . Let the other vertices be  $$B(\vec b),\;C(\vec c)\;{\text{and}}\;D(\vec d)$$ .

In the following, note carefully how the order of the cross-product is taken:

We have

\begin{align}&{{\vec n}_1} = \frac{1}{2}(\overrightarrow {BC} \times \overrightarrow {BD} ) \hfill \\\\&\quad= \frac{1}{2}\left\{ {(\vec c - \vec b) \times (\vec d - \vec b)} \right\} \hfill \\\\&\quad= \frac{1}{2}\left\{ {\vec c \times \vec d + \vec b \times \vec c + \vec d \times \vec b} \right\} \hfill \\\\& {{\vec n}_2} = \frac{1}{2}(\overrightarrow {AD} \times \overrightarrow {AC} ) \hfill \\\\&\quad = \frac{1}{2}(\vec d \times \vec c) = - \frac{1}{2}(\vec c \times \vec d) \hfill \\\\& {{\vec n}_3} = \frac{1}{2}(\overrightarrow {AB} \times \overrightarrow {AD} ) \hfill \\\\&\quad = \frac{1}{2}(\vec b \times \vec d) = - \frac{1}{2}(\vec d \times \vec b) \hfill\\ \\& {{\vec n}_4} = \frac{1}{2}(\overrightarrow {AC} \times \overrightarrow {AB} ) \hfill \\\\&\quad = \frac{1}{2}(\vec c \times \vec b) = - \frac{1}{2}(\vec b \times \vec c) \hfill \\ \end{align}

From these relations, it is clear that

${\vec n_1} + {\vec n_2} + {\vec n_3} + {\vec n_4} = \vec 0$

Example – 26

Let $$\vec a = 2\hat i + \hat k,\;\;\vec b = \hat i + \hat j + \hat k\;\;{\text{and}}\;\;\vec c = 4\hat i - 3\hat j + 7\hat k$$. Find a vector $$\vec r$$  such that $$\vec r \times \vec b = \vec c \times \vec b\;{\text{and}}\;\vec r \cdot \vec a = 0$$

Solution:                                                                                                                 $$\vec r \times \vec b = \vec c \times \vec b$$

\begin{align} & \Rightarrow \quad (\vec r - \vec c) \times \vec b = \vec 0 \hfill \\\\&\Rightarrow \quad \vec r - \vec c\,is\,{\text{ }}parallel\,{\text{ }}to\,\vec b \hfill \\ \end{align}

Thus,

\begin{align}&\vec r - \vec c = \lambda \vec b\,\,for{\text{ }}some\,\,\lambda \in \mathbb{R} \hfill \\\\& \Rightarrow \quad \vec r = \vec c + \lambda \vec b\quad\qquad\qquad....\left( 1 \right) \hfill \\ \end{align}

Now, since

\begin{align}&\qquad\vec r \cdot \vec a = 0 \hfill \\\\& \Rightarrow \quad(\vec c + \lambda \vec b) \cdot \vec a = 0 \hfill \\\\&\Rightarrow \quad \vec a \cdot \vec c + \lambda \vec a \cdot \vec b = 0 \hfill \\\\& \Rightarrow \quad \lambda = - \frac{{\vec a \cdot \vec c}}{{\vec a \cdot \vec b}} \hfill \\\\&\qquad\quad= - \frac{{8 + 7}}{{2 + 0 + 1}} \hfill \\\\&\quad\qquad = {\text{ }}-{\text{ }}5 \hfill \\ \end{align}

Using (1), we can now determine  $$\vec r$$ :

\begin{align}&\;\vec r = \vec c + \lambda \vec b \hfill \\\\&\;\;= \vec c - 5\vec b \hfill \\\\&\;\; = (4\hat i - 3\hat j + 7\hat k) - 5(\hat i + \hat j + \hat k) \hfill \\\\&\;\;= - \hat i - 8\hat j + 2\hat k \hfill \\ \end{align}

## TRY YOURSELF - III

Q. 1 If A, B, C, D be any four points in space, prove that

$\left| {\overrightarrow {AB} \times \overrightarrow {CD} + \overrightarrow {BC} \times \overrightarrow {AD} + \overrightarrow {CA} \times \overrightarrow {BD} } \right| = 4 \times \;{\text{Area}}\;(\Delta ABC)$

Q. 2 Let $$\overrightarrow {OA} = \vec a,\;\;\overrightarrow {OB} = 10\vec a + 2\vec b\;{\text{and}}\;\overrightarrow {OC} = \vec b$$ , where O is the origin. Let p denote the area of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides. Prove that p = 6q.

Q. 3 If A1, A2, A3, ... An   are the vertices of a regular plane polygon with n sides and O is its centre, prove that

$\sum\limits_{i = 1}^{n - 1} {} ({\overrightarrow {OA} _i} \times {\overrightarrow {OA} _{i + 1}}) = (n - 1)\;({\overrightarrow {OA} _1} \times {\overrightarrow {OA} _2})$

Q. 4 Let  $$\vec a = \hat i + \hat j + \hat k\;\,{\text{and}}\;\,\vec b = \hat j - \hat k$$ . Find a vector $$\vec r$$ such that  $$\vec a \times \vec r = \vec b$$ and $$\vec a \cdot \vec r = 3$$

Q. 5 Prove the sine rule for a triangle ABC with sides a, b, c :

$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}$

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