Question
Three slogans on chart papers are to be placed on a school bulletin board at the points A, Band C displaying A (Hub of Learning), B (Creating a better world for tomorrow) and C (Education comes first). The coordinates of these points are (1, 4, 2), (3, -3, -2) and (-2, 2, 6) respectively.

Based on the above information, answer the following questions.
  1. Let $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ be the position vectors of points A, B and C respectively, then $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$ is equal to:
  1. $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
  2. $2\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}}$
  3. $2\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$
  4. $2(7\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}})$
  1. Which of the following is not true?
  1. $\overline{\text{AB}}+\overline{\text{BC}}+\overline{\text{CA}}=\vec{0}$
  2. $\overline{\text{AB}}+\overline{\text{BC}}-\overline{\text{AC}}=\vec{0}$
  3. $\overline{\text{AB}}+\overline{\text{BC}}-\overline{\text{CA}}=\vec{0}$
  4. $\overline{\text{AB}}-\overline{\text{CB}}+\overline{\text{CA}}=\vec{0}$
  1. Area of $\triangle\text{ABC}$ is:
  1. 19 sq. units
  2. $\sqrt{1937}\text{sq}.\text{units}$
  3. $\frac{1}{2}\sqrt{1937}\text{sq}.\text{units}$
  4. $\sqrt{1837}\text{sq}.\text{units}$
  1. Suppose, if the given slogans are to be placed on a straight line, then the value of $|\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}|$ will be equal to:
  1. -1
  2. -2
  3. 2
  4. 0
  1. If $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}},$ then unit vector in the direction of vector $\vec{\text{a}}$ is:
  1. $\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}$
  2. $\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}+\frac{6}{7}\hat{\text{k}}$
  3. $\frac{3}{7}\hat{\text{i}}+\frac{2}{7}\hat{\text{j}}+\frac{6}{7}\hat{\text{k}}$
  4. None of these

Answer

  1. (a) $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
Solution:

$\vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},\vec{\text{ b}}=3\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}$

And $\vec{\text{c}}=2\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$

$\therefore\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
  1. (c) $\overline{\text{AB}}+\overline{\text{BC}}-\overline{\text{CA}}=\vec{0}$
Solution:

Using triangle law of addition in $\triangle\text{ABC},$ we get $\overline{\text{AB}}+\overline{\text{BC}}-\overline{\text{CA}}=\vec{0}$ which can be rewritten as,

$\overline{\text{AB}}+\overline{\text{BC}}-\overline{\text{CA}}=\vec{0}$ or $\overline{\text{AB}}-\overline{\text{CB}}+\overline{\text{CA}}=\vec{0}$
  1. (c) $\frac{1}{2}\sqrt{1937}\text{sq}.\text{units}$
Solution:

We have, A(1 ,4, 2), B(3, -3, -2) and C(-2, 2, 6)

Now, $\overline{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}=2\hat{\text{i}}-7\hat{\text{j}}-4\hat{\text{k}}$

And $\overline{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}=-3\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$

$\therefore\overline{\text{AB}}\times\overline{\text{AC}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-7&-4\\-3&-2&4\end{vmatrix}$

$=\hat{\text{i}}(-28-8)-\hat{\text{j}}(8-12)+\hat{\text{k}}(-4-21)$

$=-36\hat{\text{i}}+4\hat{\text{j}}-25\hat{\text{k}}$

Now, $|\overline{\text{AB}}\times\overline{\text{AC}}|=\sqrt{(-36)^2+4^2+(-25)^2}$

$=\sqrt{1296+16+625}=\sqrt{1937}$

$\therefore$ Area of $\triangle\text{ABC}=\frac{1}{2}|\overline{\text{AB}}\times\overline{\text{AC}}|$

$=\frac{1}{2}\sqrt{1937}\text{sq}.\text{units}.$
  1. (d) 0
Solution:

If the given points lie on the straight line, then the points will be collinear and so area of $\triangle\text{ABC}=0.$

$\Rightarrow|\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}|=0$

[$\because$ If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of the three vertices A, B and C of $\triangle\text{ABC},$ then area of triangle

$=\frac{1}{2}|\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}|]$
  1. (b) $\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}+\frac{6}{7}\hat{\text{k}}$
Solution:

Here, $|\vec{\text{a}}|=\sqrt{2^2+3^2+6^2}=\sqrt{4+6+36}$

$=\sqrt{49}=7$

$\therefore$ Unit vector in the direction of vector $\vec{\text{a}}$ is

$\hat{\text{a}}=\frac{2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}}{7}$

$=\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}+\frac{6}{7}\hat{\text{k}}$

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Image
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  2. $21$
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  4. $\frac{4}{3}$
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  2. $20$
  3. $\frac{4}{3}$
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Image

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Then which of the following point lie in its feasible region.
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