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+4 votes
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in Mathematics by (130k points)
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The two adjacent sides of a parallelogram are 2i-4j+5k and i-2j-3k Find the unit vector parallel to its diagonal. Also, find its area.

2 Answers

+2 votes
by (17.0k points)
selected by
 
Best answer

Given, two adjacent sides of the parallelogram 

\(2\hat i - 4\hat j + 5\hat k \) and \(\hat i - 2\hat j - 3\hat k\)

The diagonal will be the resultant of these two vectors.

So resultant R:

\(\vec R = 2\hat i - 4\hat j + 5\hat k + \hat i -2 \hat j - 3 \hat k\)

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

Now unit vector in direction of R

\(\vec u = \frac{3\hat i - 6\hat j + 2\hat k}{\sqrt{3^2 + (-6)^2 + 2^2}}\)

\(= \frac{3\hat i - 6\hat j + 2 \hat k}{\sqrt{49}}\)

\(= \frac{3\hat i - 6\hat j + 2\hat k}{7}\)

Hence unit vector along the diagonal of the parallelogram

\(\vec u= \frac 37 \hat i - \frac 67 \hat j + \frac 27 \hat k\)

Now,

Area of parallelogram

\(A = (2\hat i - 4\hat j + 5\hat k) \times (\hat i - 2 \hat j - 3 \hat k)\)

\(A = \begin{vmatrix}\hat i &\hat j & \hat k\\2&-4&5\\1&-2&-3\end{vmatrix}\)

\(= |\hat i (12 + 10) - \hat j (-6 - 5) + \hat k(-4 + 4)|\)

\(= |22\hat i + 11\hat j|\)

\(A = \sqrt{22^2 + 11^2 }\)

\(= 11\sqrt 5\)

Hence the area of the parallelogram is 11√5.

+4 votes
by (93.7k points)

Thus, the unit vector parallel to the diagonal is

Hence, the area of the parallelogram is 11√5 square units.

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