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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA4 Marks
Question
A barge is pulled into harbour by two tug boats as shown in the figure.
Based on the above information, answer the following questions.
  1. Position vector of A is:
  1. $4\hat{\text{i}}+2\hat{\text{j}}$
  2. $4\hat{\text{i}}+10\hat{\text{j}}$
  3. $4\hat{\text{i}}-10\hat{\text{j}}$
  4. $4\hat{\text{i}}-2\hat{\text{j}}$
  1. Position vector of B is:
  1. $4\hat{\text{i}}+4\hat{\text{j}}$
  2. $6\hat{\text{i}}+6\hat{\text{j}}$
  3. $9\hat{\text{i}}+7\hat{\text{j}}$
  4. $3\hat{\text{i}}+3\hat{\text{j}}$
  1. Find the vector $\overline{\text{AC}}$ in terms of $\hat{\text{i}},\hat{\text{j}}.$
  1. $8\hat{\text{j}}$
  2. $-8\hat{\text{j}}$
  3. $8\hat{\text{i}}$
  4. None of these
  1. If $\vec{\text{A}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},$ then its unit vector is:
  1. $\frac{\hat{\text{i}}}{\sqrt{14}}+\frac{2\hat{\text{j}}}{\sqrt{14}}+\frac{3\hat{\text{k}}}{\sqrt{14}}$
  2. $\frac{3\hat{\text{i}}}{\sqrt{14}}+\frac{2\hat{\text{j}}}{\sqrt{14}}+\frac{\hat{\text{k}}}{\sqrt{14}}$
  3. $\frac{2\hat{\text{i}}}{\sqrt{14}}+\frac{3\hat{\text{j}}}{\sqrt{14}}+\frac{\hat{\text{k}}}{\sqrt{14}}$
  4. None of these
  1. If $\vec{\text{A}}=4\hat{\text{i}}+3\hat{\text{j}}$ and $\vec{\text{B}}=3\hat{\text{i}}+4\hat{\text{j}},$ then $|\vec{\text{A}}|+|\vec{\text{B}}|=$
  1. 12
  2. 13
  3. 14
  4. 10
✓

Answer

  1. (b) $4\hat{\text{i}}+10\hat{\text{j}}$
Solution:

Here, (4, 10) are the coordinates of A.

$\therefore\text{P.V of }\text{A}=4\hat{\text{i}}+10\hat{\text{j}}$
  1. (c) $9\hat{\text{i}}+7\hat{\text{j}}$
Solution:

Here, (9, 7) are the coordinates of B.

$\therefore\text{P.V of }\text{B}=9\hat{\text{i}}+7\hat{\text{j}}$
  1. (b) $-8\hat{\text{j}}$
Solution:

Here, P.V. of $\text{A}=4\hat{\text{i}}+10\hat{\text{j}}$ and P.V. of

$\text{C}=4\hat{\text{i}}+2\hat{\text{j}}$

$\therefore\overline{\text{AC}}=(4-4)\hat{\text{i}}+(2-10)\hat{\text{j}}=-8\hat{\text{j}}$
  1. (a) $\frac{\hat{\text{i}}}{\sqrt{14}}+\frac{2\hat{\text{j}}}{\sqrt{14}}+\frac{3\hat{\text{k}}}{\sqrt{14}}$
Solution:

Here, $\vec{\text{A}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$

$\therefore|\vec{\text{A}}|=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

$\therefore\vec{\text{A}}=\frac{\vec{\text{A}}}{|\vec{\text{A}}|}=\frac{\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{14}}$

$=\frac{1}{\sqrt{14}}\hat{\text{i}}+\frac{2}{\sqrt{14}}\hat{\text{j}}+\frac{3}{\sqrt{14}}\hat{\text{k}}$
  1. (d) 10
Solution:

We have, $\vec{\text{A}}=4\hat{\text{i}}+3\hat{\text{j}}$ and $\vec{\text{B}}=3\hat{\text{i}}+4\hat{\text{j}}$

$\therefore|\vec{\text{A}}|=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

And $|\vec{\text{B}}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

Thus, $|\vec{\text{A}}|+|\vec{\text{B}}|=5+5=10.$

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