# 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.

Answer

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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.

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.

Last updated date: 20th Sep 2023

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