यदि \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=2\hat{i}-\hat{j}+3\hat{k}\) और \(\vec{c}=\hat{i}-2\hat{j}+\hat{k}\), और सदिश \(2\vec{a}-\vec{b}+3\vec{c}\) के समानांतर एक इकाई सदिश कितना होगा?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 3)

\(\frac{9}{{\sqrt {22} }}\hat{i}-\frac{3}{{\sqrt {22} }}\hat{j}+\frac{1}{{\sqrt {22} }}\hat{k}\)
\(\frac{2}{{\sqrt {22} }}\hat{i}-\frac{3}{{\sqrt {22} }}\hat{j}+\frac{2}{{\sqrt {22} }}\hat{k}\)
\(\frac{3}{{\sqrt {22} }}\hat{i}-\frac{3}{{\sqrt {22} }}\hat{j}+\frac{3}{{\sqrt {22} }}\hat{k}\)
\(\frac{3}{{\sqrt {22} }}\hat{i}-\frac{3}{{\sqrt {22} }}\hat{j}+\frac{2}{{\sqrt {22} }}\hat{k}\)

Answer 1

दिया गया है:

\(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)

\(\vec{b}=2\hat{i}-\hat{j}+3\hat{k}\)

\(\vec{c}=\hat{i}-2\hat{j}+\hat{k}\)

गणना:

सदिश की दिशा में इकाई सदिश, \(\vec r = 2\vec{a}-\vec{b}+3\vec{c}\)

⇒ \(\vec r = 2(\hat i + \hat j + \hat k) - (2\hat i - \hat j + 3\hat k) + 3(\hat i - 2\hat j + \hat k)\)

⇒ \(\vec r = 2 \hat i + 2\hat j + 2\hat k - 2 \hat i + \hat j - 3\hat k + 3 \hat i - 6 \hat j + 3 \hat k\)

⇒ \(\vec r = (2 - 2 + 3) \hat i + (2 + 1 - 6) \hat j + (2 - 3 + 3) \hat k\)

⇒ \(\vec r = 3 \hat i - 3 \hat j + 2 \hat k\)

परिमाण or \(\vec r\) = \(\sqrt{3^2 + (-3)^2 + 2^2}\)

⇒ \(|\vec r|\) = \(\sqrt{9+ 9+ 4}\) = \(\sqrt{22}\)

\(\vec r = \frac{1}{| \vec r|}\times \vec r\) की दिशा में इकाई सदिश

⇒ \(\vec r = \frac{1}{\sqrt{22}} \times [3 \hat i - 3 \hat j + 2 \hat k]\)

⇒ \(\vec r = \frac{3}{\sqrt{22}} \hat i - \frac{3}{\sqrt{22}} \hat j + \frac{2}{\sqrt{22}} \hat k\)

∴ \(\frac{3}{\sqrt{22}} \hat i - \frac{3}{\sqrt{22}} \hat j + \frac{2}{\sqrt{22}} \hat k\) अभीष्ट सदिश है