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# Write the sum of intercepts cut off by the place $\overrightarrow{r}.\left( 2\widehat{i}+\widehat{j}-\widehat{k} \right)-5=0$ on the three axes.

Last updated date: 14th Jul 2024
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Hint: Here, first off substitute $r=x\widehat{i}+y\widehat{j}+z\widehat{k}$ to find the equation of plane in ax + by + cz + d = 0 form. Then we know that plane in intercept form is $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ where a, b and c are intercepts on x, y and z axes. So, convert the given plane into the plane in the intercept form to get the sum of the intercepts.

Complete step by step solution:
Here we have to find the sum of the intercepts cut off by the plane $\overrightarrow{r}.\left( 2\widehat{i}+\widehat{j}-\widehat{k} \right)-5=0$ on three axes.
Let us first consider the equation of the plane given in the question.
$P=\overrightarrow{r}.\left( 2\widehat{i}+\widehat{j}-\widehat{k} \right)-5=0$
We know that $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$. By substituting the value of $\overrightarrow{r}$ in the above equation, we get,
$P:\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\left( 2\widehat{i}+\widehat{j}-\widehat{k} \right)-5=0$
We know that $\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( m\widehat{i}+n\widehat{j}+q\widehat{k} \right)=am+bn+cq$
By using this, we get,
$P:2x+y-z-5=0$
By dividing the whole equation by 5, we get,
$P:\dfrac{2x}{5}+\dfrac{y}{5}-\dfrac{z}{5}=1$
We can also write the above equation as,
$P:\dfrac{x}{\dfrac{5}{2}}+\dfrac{y}{5}+\dfrac{z}{\left( -5 \right)}=1....\left( i \right)$
We know that the equation of a plane in intercept form is given by $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ where a, b and c are the intercepts cut off by plane in x, y and z axes respectively.
By comparing equation (i) with equation of plane in the intercept form that is $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$
We get,
\begin{align} & a=\dfrac{5}{2} \\ & b=5 \\ & c=-5 \\ \end{align}
Hence, we get,
Intercept cut off by the place on the x axis $=\dfrac{5}{2}$
Intercept cut off by the plane on y-axis = 5
Intercept cut off by the place on z-axis = -5
Hence, we get the sum of the intercept cut off by plane on all axes
$=\dfrac{5}{2}+5-5$
$=\dfrac{5}{2}=2.5$
Therefore, we have formed the sum of the intercepts as $\dfrac{5}{2}=2.5$

Note: Students must note that the negative intercept signifies that the intercept cut off by the given plane is in the negative axis. For example, if we get ‘-a’ as an intercept on the x-axis, that means the intercept cut off by plane is ‘a’ on the negative x-axis or on the left side of the origin. Also, students are advised to always first substitute $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$ and then solve the questions related to the plane to get the answers easily.