Question

The two adjacent sides of a parallelogram are 2i-4j-5k and 2i+2j+3k. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.

Answer 1



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Secondary SchoolMath13 points

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

Ask for details Follow Report byArtitanwar8009 13.03.2018

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The unit vector to the diagonal is (3i - 6j + 2k) / 7 and the area of the parallelogram is 11 (5)^0.5

The diagonal of a parallelogram whose adjacent sides a and b are given, is calculated using the formula:

a + b (where both a and b should be in vector notation)
a + b = (i-2j-3k) + (2i-4j+5k)
a + b = 3i - 6j + 2k
Magnitude of a + b is 7
Hence unit vector to the diagonal is (3i - 6j + 2k) / 7

Area of parallelogram is given by formula:

A = 0.5 [a x b]
A = 0.5 [22i + 11j]
A =  11 (5)^0.5

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Answer 2

Answer:

Step-by-step explanation:

let consider the two side of the parallelogram are $\vec{P}$ and $\vec{Q}$

Given,  

$\vec{P}=2\hat{i}-4\hat{j}-5\hat{k}$

$\vec{Q}=2\hat{i}+2\hat{j}+3\hat{k}$

Let the two diagonals of the parallelogram are $\vec{R}$ and $\vec{S}$

Hence,

$\vec{R}=\vec{P}+\vec{Q}\\\\\Rightarrow{\vec{R}}=(2\hat{i}-4\hat{j}-5\hat{k})+(2\hat{i}+2\hat{j}+3\hat{k})\\\\\Rightarrow{\vec{R}}=4\hat{i}-2\hat{j}-2\hat{k}$

$\vec{S}=\vec{Q}+\vec{P}\\\\\Rightarrow{\vec{S}}=(2\hat{i}+2\hat{j}+3\hat{k})-(2\hat{i}-4\hat{j}-5\hat{k})\\\\\Rightarrow{\vec{S}}=6\hat{j}+8\hat{k}$

Therefore the magnitude of $\left|\vec{R}\right|=\sqrt{(4)^{2}+(-2)^{2}+(-2)^{2}}=\sqrt{16+4+4}=\sqrt{24}$

Therefore the magnitude of $\left|\vec{S}\right|=\sqrt{(6)^{2}+(8)^{2}}=\sqrt{36+84}=\sqrt{100}=10$

The unit vector in the direction of $\vec{R}=\frac{1}{\sqrt{24}}\times(4\hat{i}-2\hat{j}-2\hat{k})$

The unit vector in the direction of $\vec{S}=\frac{1}{10}\times(6\hat{j}+8\hat{k})$

Therefore the unit vectors parallel to its diagonals are

$\vec{R}=\frac{1}{\sqrt{24}}\times(4\hat{i}-2\hat{j}-2\hat{k})$

and  

$\vec{S}=\frac{1}{10}\times(6\hat{j}+8\hat{k})$

We know that area of a parallelogram = $\left|\vec{P}\times\vec{Q}\right|$

So,

$\vec{P}\times\vec{Q}=\begin{bmatrix}\hat{i}&\hat{j}&\hat{k}\\2&(-4)&(-5)\\2&2&3\end{bmatrix}$

$\vec{P}\times\vec{Q}=\hat{i}[(3\times(-4))-(2\times(-5))]-\hat{j}[(2\times3)-(2\times(-5))+\hat{k}[(2\times2)-((-4)\times2)]$

$\vec{P}\times\vec{Q}=\hat{i}[-12+10]-\hat{j}[6+10]+\hat{k}[4+8]=-2\hat{i}-16\hat{j}+12\hat{k}$

So,  

$\left|\vec{P}\times\vec{Q}\right|=\sqrt{(-2)^{2}+(-16)^{2}+(12)^{2}}\\\\\Rightarrow\left|\vec{P}\times\vec{Q}\right|=\sqrt{4+256+144}=\sqrt{404}=2\sqrt{101}$