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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Vector Algebra

BITSAT 2010 · Mathematics · Vector Algebra

With respect to a rectangular cartesian coordinate system, three...

With respect to a rectangular cartesian coordinate system, three vectors are expressed as : (\vec{a}=4 \hat{i}-\hat{j}, \vec{b}=-3 \hat{i}+2 \hat{j}) and (\vec{c}=-\hat{k} \quad) where (\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}}) are unit vectors, along the XX, YY and ZZ-axis respectively. The unit vector (\hat{r}) along the direction of sum of these vector is -

Options

  1. A(\hat{\mathrm{r}}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}))
  2. B(\hat{\mathrm{r}}=\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}))
  3. C(\hat{\mathrm{i}}=\frac{1}{3}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}))
  4. D(\hat{\mathrm{r}}=\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}))

Correct answer

A. (\hat{\mathrm{r}}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}))

Step-by-step solution

(r→=a→+b→+c→=4i^−j^−3i^+2j^−k^=i^+j^−k^r^=r→∣r∣=i^+j^−k^12+12+(−1)2=i^+j^−k^3\begin{aligned} & \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\mathrm{r}|}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{1^2+1^2+(-1)^2}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}} \end{aligned})
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