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Question: Reduce the equation \[\overrightarrow{r}.\left( \overrightarrow{i}-2\overrightarrow{j}+2\overrightar...

Reduce the equation r.(i2j+2k)  6 = 0\overrightarrow{r}.\left( \overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k} \right)\text{ }-\text{ }6\text{ }=\text{ }0 to normal form and hence find the length of perpendicular from the origin to the plane.

Explanation

Solution

Hint: Convert the given equation in the form of r.n=p\overrightarrow{r}.\overrightarrow{n}=\text{p} and then use the formula of given by r.nn=pn\overrightarrow{r}.\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}=\dfrac{\text{p}}{\left| \overrightarrow{n} \right|} to get the normal form of the given equation. Also, the value of pn\dfrac{\text{p}}{\left| \overrightarrow{n} \right|} is the perpendicular distance from the origin.

Complete step-by-step answer:
To solve the above problem we will write the given equation first, therefore,
r.(i2j+2k)  6 = 0\overrightarrow{r}.\left( \overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k} \right)\text{ }-\text{ }6\text{ }=\text{ }0
If we shift -6 on the right hand side of the equation we will get,
r.(i2j+2k) = 6\therefore \overrightarrow{r}.\left( \overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k} \right)\text{ }=\text{ 6}
If we compare the above equation with r.n=p\overrightarrow{r}.\overrightarrow{n}=\text{p} we will get,
n=(i2j+2k)\overrightarrow{n}=\left( \overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k} \right) and p = 6 …………………………………………………………………….. (1)
Now we have to find out n\left| \overrightarrow{n} \right| to reduce the equation in to normal form and for that we should know the formula given below,
Formula:
If n=ai+bj+ck\overrightarrow{n}=a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k} then n\left| \overrightarrow{n} \right| is given by n=a2+b2+c2\left| \overrightarrow{n} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}
By using the formula and the values of equation (1) we will find the value of n\left| \overrightarrow{n} \right| for given equation as follows,
n=12+(2)2+(2)2\therefore \left| \overrightarrow{n} \right|=\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( 2 \right)}^{2}}}
n=1+4+4\therefore \left| \overrightarrow{n} \right|=\sqrt{1+4+4}
n=9\therefore \left| \overrightarrow{n} \right|=\sqrt{9}
As we know that the square root of 9 is 3 therefore above equation will become,
n=3\therefore \left| \overrightarrow{n} \right|=3 …………………………………………………. (2)
Now, to reduce the equation in the normal form we should know the formula given below,
Formula:
The normal form of the equation r.n=p\overrightarrow{r}.\overrightarrow{n}=\text{p} is given by r.nn=pn\overrightarrow{r}.\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}=\dfrac{\text{p}}{\left| \overrightarrow{n} \right|}.
If we put the value of equation (1) and equation (2) in the above formula we will get,
r.(i2j+2k)3=63\therefore \overrightarrow{r}.\dfrac{\left( \overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k} \right)}{3}=\dfrac{6}{3}
If we simplify the above equation we will get,
r.(i32j3+2k3)=2\therefore \overrightarrow{r}.\left( \dfrac{\overrightarrow{i}}{3}-\dfrac{2\overrightarrow{j}}{3}+\dfrac{2\overrightarrow{k}}{3} \right)=2 ……………………………………………………… (4)
Therefore the normal form of the equation is r.(i2j+2k)  6 = 0\overrightarrow{r}.\left( \overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k} \right)\text{ }-\text{ }6\text{ }=\text{ }0 is r.(13i23j+23k)=2\overrightarrow{r}.\left( \dfrac{1}{3}\overrightarrow{i}-\dfrac{2}{3}\overrightarrow{j}+\dfrac{2}{3}\overrightarrow{k} \right)=2.
As we know that nn=n ^\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}=\overset{\hat{\ }}{\mathop{n}}\, and therefore the normal form of the equation can also be written as r.n ^=d\overrightarrow{r}.\overset{\hat{\ }}{\mathop{n}}\,=d where ‘d’ is the distance of the plane from the origin which is the required perpendicular distance from the origin.
As the equation (4) is of the form r.n ^=d\overrightarrow{r}.\overset{\hat{\ }}{\mathop{n}}\,=d therefore,
d = 2
Therefore the length of perpendicular from the origin is equal to 2.

Note: Always remember that the equation r.n=p\overrightarrow{r}.\overrightarrow{n}=\text{p} and r.n ^=d\overrightarrow{r}.\overset{\hat{\ }}{\mathop{n}}\,=d are not same as many students write the value of p in place of d which results in incorrect answer.