Question

Find the equation of the three planes which are parallel to the co-ordinate axes and which pass through the points (3,1,1) and (1,-2,3) ???​

Answer 1

Answer: 1) 2y + 3z = 5

              2) x + z = 4

              3) 3x - 2y = 7

Step-by-step explanation:  

To find that equation of plane,  

We need to find a point P which lies on the plane and a another vector which is perpendicular to the required plane.  

Equation of plane passing through a point a and perpendicular to vector n is given by,  

$(\vec{r}-\vec{P})\cdot\vec{n}=0$  

$\vec{OP}=3\hat{i}+1\hat{j}+1\hat{k}\\\;\\\vec{OQ}=1\hat{i}-2\hat{j}+3\hat{k}\\\;\\\vec{PQ}=\vec{OQ}-\vec{OP}\\\;\\\vec{PQ}=(1\hat{i}-2\hat{j}+3\hat{k})-(3\hat{i}+1\hat{j}+1\hat{k})\\\;\\\vec{PQ}=-2\hat{i}-3\hat{j}+2\hat{k}$

Case: 1 Equation of plane Parallel to x-axis  

If the plane is parallel to x-axis then Normal of the plane should be perpendicular to x - axis.  

Let a unit vector $\hat{i}$ .  

Now that normal of the plane will be such that, which is perpendicular to both $\vec{PQ}$ and $\hat{i}$  

We can say that,  

$\vec{n}=\hat{i}\;\times\;(-2\hat{i}-3\hat{j}+2\hat{k})\\\;\\\vec{n}=-2(\hat{i}\times\hat{i})-3(\hat{i}\times\hat{j})+2(\hat{i}\times\hat{k})\\\;\\\vec{n}=\vec{0}-3\hat{k}-2\hat{j}\\\;\\\vec{n}=0\hat{i}-2\hat{j}-3\hat{k}$  

Now,  

Required Equation of Plane will be,  

$(\vec{r}-\vec{P})\cdot\vec{n}=0$

$(\vec{r}-(3\hat{i}+1\hat{j}+1\hat{k}))\cdot(0\hat{i}-2\hat{j}-3\hat{k})=0\\\;\\\text{Putting}\;\;\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\;\\((x-3)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k})\cdot(0\hat{i}-2\hat{j}-3\hat{k})=0\\\;\\0(x-3)-2(y-1)-3(z-1)=0\\\;\\-2y+2-3z+3=0\\\;\\-2y-3z+5=0\\\;\\2y+3z-5=0\\\;\\2y+3z=5$  

This is the required equation of the plane which is parallel to x-axis and passing through the points (3,1,1) and (1,-2,3).  

Case: 2 Equation of plane Parallel to y-axis  

If the plane is parallel to y-axis then Normal of the plane should be perpendicular to y - axis.  

Let a unit vector $\hat{j}$ .  

Now that normal of the plane will be such that, which is perpendicular to both $\vec{PQ}$ and $\hat{j}$  

We can say that,  

$\vec{n}=\hat{j}\;\times\;(-2\hat{i}-3\hat{j}+2\hat{k})\\\;\\\vec{n}=-2(\hat{j}\times\hat{i})-3(\hat{j}\times\hat{j})+2(\hat{j}\times\hat{k})\\\;\\\vec{n}=2\hat{k}+\vec{0}+2\hat{i}\\\;\\\vec{n}=2\hat{i}-0\hat{j}+2\hat{k}$

Now,  

Required Equation of Plane will be,  

$(\vec{r}-\vec{P})\cdot\vec{n}=0\\\;\\(\vec{r}-(3\hat{i}+1\hat{j}+1\hat{k}))\cdot(2\hat{i}-0\hat{j}+2\hat{k})=0\\\;\\\text{Putting}\;\;\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\;\\((x-3)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k})\cdot(2\hat{i}-0\hat{j}+2\hat{k})=0\\\;\\2(x-3)-0(y-1)+2(z-1)=0\\\;\\2x-6+2z-2=0\\\;\\2x+2z-8=0\\\;\\2x+2z=8\\\;\\x+z=4$

This is the required equation of the plane which is parallel to y-axis and passing through the points (3,1,1) and (1,-2,3).  

Case: 3 Equation of plane Parallel to z-axis  

If the plane is parallel to x-axis then Normal of the plane should be perpendicular to z - axis.  

Let a unit vector $\hat{k}$ .  

Now that normal of the plane will be such that, which is perpendicular to both $\vec{PQ}$ and $\hat{k}$  

We can say that,  

$\vec{n}=\hat{k}\;\times\;(-2\hat{i}-3\hat{j}+2\hat{k})\\\;\\\vec{n}=-2(\hat{k}\times\hat{i})-3(\hat{k}\times\hat{j})+2(\hat{k}\times\hat{k})\\\;\\\vec{n}=-2\hat{j}+3\hat{i}+\vec{0}\\\;\\\vec{n}=-2\hat{j}+3\hat{i}+0\hat{k}\\\;\\\vec{n}=3\hat{i}-2\hat{j}+0\hat{k}$

Now,  

Required Equation of Plane will be,  

$(\vec{r}-\vec{P})\cdot\vec{n}=0\\\;\\(\vec{r}-(3\hat{i}+1\hat{j}+1\hat{k}))\cdot(3\hat{i}-2\hat{j}+0\hat{k})=0\\\;\\\text{Putting}\;\;\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\;\\((x-3)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k})\cdot(3\hat{i}-2\hat{j}+0\hat{k})=0\\\;\\3(x-3)-2(y-1)+0(z-1)=0\\\;\\3x-9-2y+2=0\\\;\\3x-2y-7=0\\\;\\3x-2y=7$

This is the required equation of the plane which is parallel to z-axis and passing through the points (3,1,1) and (1,-2,3).

Answer 2

Answer:

Answer: 1) 2y + 3z = 5

2) x + z = 4

3) 3x - 2y = 7

Step-by-step explanation:

To find that equation of plane,

We need to find a point P which lies on the plane and a another vector which is perpendicular to the required plane.

Equation of plane passing through a point a and perpendicular to vector n is given by,

(\vec{r}-\vec{P})\cdot\vec{n}=0(

r

−

P

)⋅

n

=0

\begin{gathered}\vec{OP}=3\hat{i}+1\hat{j}+1\hat{k}\\\;\\\vec{OQ}=1\hat{i}-2\hat{j}+3\hat{k}\\\;\\\vec{PQ}=\vec{OQ}-\vec{OP}\\\;\\\vec{PQ}=(1\hat{i}-2\hat{j}+3\hat{k})-(3\hat{i}+1\hat{j}+1\hat{k})\\\;\\\vec{PQ}=-2\hat{i}-3\hat{j}+2\hat{k}\end{gathered}

OP

=3

i

^

+1

j

^

+1

k

^

OQ

=1

i

^

−2

j

^

+3

k

^

PQ

=

OQ

−

OP

PQ

=(1

i

^

−2

j

^

+3

k

^

)−(3

i

^

+1

j

^

+1

k

^

)

PQ

=−2

i

^

−3

j

^

+2

k

^

Case: 1 Equation of plane Parallel to x-axis

If the plane is parallel to x-axis then Normal of the plane should be perpendicular to x - axis.

Let a unit vector \hat{i}

i

^

.

Now that normal of the plane will be such that, which is perpendicular to both \vec{PQ}

PQ

and \hat{i}

i

^

We can say that,

\begin{gathered}\vec{n}=\hat{i}\;\times\;(-2\hat{i}-3\hat{j}+2\hat{k})\\\;\\\vec{n}=-2(\hat{i}\times\hat{i})-3(\hat{i}\times\hat{j})+2(\hat{i}\times\hat{k})\\\;\\\vec{n}=\vec{0}-3\hat{k}-2\hat{j}\\\;\\\vec{n}=0\hat{i}-2\hat{j}-3\hat{k}\end{gathered}

n

=

i

^

×(−2

i

^

−3

j

^

+2

k

^

)

n

=−2(

i

^

×

i

^

)−3(

i

^

×

j

^

)+2(

i

^

×

k

^

)

n

=

0

−3

k

^

−2

j

^

n

=0

i

^

−2

j

^

−3

k

^

Now,

Required Equation of Plane will be,

(\vec{r}-\vec{P})\cdot\vec{n}=0(

r

−

P

)⋅

n

=0

\begin{gathered}(\vec{r}-(3\hat{i}+1\hat{j}+1\hat{k}))\cdot(0\hat{i}-2\hat{j}-3\hat{k})=0\\\;\\\text{Putting}\;\;\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\;\\((x-3)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k})\cdot(0\hat{i}-2\hat{j}-3\hat{k})=0\\\;\\0(x-3)-2(y-1)-3(z-1)=0\\\;\\-2y+2-3z+3=0\\\;\\-2y-3z+5=0\\\;\\2y+3z-5=0\\\;\\2y+3z=5\end{gathered}

(

r

−(3

i

^

+1

j

^

+1

k

^

))⋅(0

i

^

−2

j

^

−3

k

^

)=0

Putting

r

=x

i

^

+y

j

^

+z

k

^

((x−3)

i

^

+(y−1)

j

^

+(z−1)

k

^

)⋅(0

i

^

−2

j

^

−3

k

^

)=0

0(x−3)−2(y−1)−3(z−1)=0

−2y+2−3z+3=0

−2y−3z+5=0

2y+3z−5=0

2y+3z=5

This is the required equation of the plane which is parallel to x-axis and passing through the points (3,1,1) and (1,-2,3).

Case: 2 Equation of plane Parallel to y-axis

If the plane is parallel to y-axis then Normal of the plane should be perpendicular to y - axis.

Let a unit vector \hat{j}

j

^

.

Now that normal of the plane will be such that, which is perpendicular to both \vec{PQ}

PQ

and \hat{j}

j

^

We can say that,

\begin{gathered}\vec{n}=\hat{j}\;\times\;(-2\hat{i}-3\hat{j}+2\hat{k})\\\;\\\vec{n}=-2(\hat{j}\times\hat{i})-3(\hat{j}\times\hat{j})+2(\hat{j}\times\hat{k})\\\;\\\vec{n}=2\hat{k}+\vec{0}+2\hat{i}\\\;\\\vec{n}=2\hat{i}-0\hat{j}+2\hat{k}\end{gathered}

n

=

j

^

×(−2

i

^

−3

j

^

+2

k

^

)

n

=−2(

j

^

×

i

^

)−3(

j

^

×

j

^

)+2(

j

^

×

k

^

)

n

=2

k

^

+

0

+2

i

^

n

=2

i

^

−0

j

^

+2

k

^

Now,

Required Equation of Plane will be,

\begin{gathered}(\vec{r}-\vec{P})\cdot\vec{n}=0\\\;\\(\vec{r}-(3\hat{i}+1\hat{j}+1\hat{k}))\cdot(2\hat{i}-0\hat{j}+2\hat{k})=0\\\;\\\text{Putting}\;\;\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\;\\((x-3)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k})\cdot(2\hat{i}-0\hat{j}+2\hat{k})=0\\\;\\2(x-3)-0(y-1)+2(z-1)=0\\\;\\2x-6+2z-2=0\\\;\\2x+2z-8=0\\\;\\2x+2z=8\\\;\\x+z=4\end{gathered}

(

r

−

P

)⋅

n

=0

(

r

−(3

i

^

+1

j

^

+1

k

^

))⋅(2

i

^

−0

j

^

+2

k

^

)=0

Putting

r

=x

i

^

+y

j

^

+z

k

^

((x−3)

i

^

+(y−1)

j

^

+(z−1)

k

^

)⋅(2

i

^

−0

j

^

+2

k

^

)=0

2(x−3)−0(y−1)+2(z−1)=0

2x−6+2z−2=0

2x+2z−8=0

2x+2z=8

x+z=4

This is the required equation of the plane which is parallel to y-axis and passing through the points (3,1,1) and (1,-2,3).

Case: 3 Equation of plane Parallel to z-axis

If the plane is parallel to x-axis then Normal of the plane should be perpendicular to z - axis.

Let a unit vector \hat{k}

k

^

.

Now that normal of the plane will be such that, which is perpendicular to both \vec{PQ}

PQ

and \hat{k}

k

^

We can say that,

\begin{gathered}\vec{n}=\hat{k}\;\times\;(-2\hat{i}-3\hat{j}+2\hat{k})\\\;\\\vec{n}=-2(\hat{k}\times\hat{i})-3(\hat{k}\times\hat{j})+2(\hat{k}\times\hat{k})\\\;\\\vec{n}=-2\hat{j}+3\hat{i}+\vec{0}\\\;\\\vec{n}=-2\hat{j}+3\hat{i}+0\hat{k}\\\;\\\vec{n}=3\hat{i}-2\hat{j}+0\hat{k}\end{gathered}

n

=

k

^

×(−2

i

^

−3

j

^

+2

k

^

)

n

=−2(

k

^

×

i

^

)−3(

k

^

×

j

^

)+2(

k

^

×

k

^

)

n

=−2

j

^

+3

i

^

+

0

n

=−2

j

^

+3

i

^

+0

k

^

n

=3

i

^

−2

j

^

+0

k

^

Now,

Required Equation of Plane will be,

\begin{gathered}(\vec{r}-\vec{P})\cdot\vec{n}=0\\\;\\(\vec{r}-(3\hat{i}+1\hat{j}+1\hat{k}))\cdot(3\hat{i}-2\hat{j}+0\hat{k})=0\\\;\\\text{Putting}\;\;\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\;\\((x-3)\hat{i}+(y-1)\hat{j}+(z-1)\hat{k})\cdot(3\hat{i}-2\hat{j}+0\hat{k})=0\\\;\\3(x-3)-2(y-1)+0(z-1)=0\\\;\\3x-9-2y+2=0\\\;\\3x-2y-7=0\\\;\\3x-2y=7\end{gathered}

(

r

−

P

)⋅

n

=0

(

r

−(3

i

^

+1

j

^

+1

k

^

))⋅(3

i

^

−2

j

^

+0

k

^

)=0

Putting

r

=x

i

^

+y

j

^

+z

k

^

((x−3)

i

^

+(y−1)

j

^

+(z−1)

k

^

)⋅(3

i

^

−2

j

^

+0

k

^

)=0

3(x−3)−2(y−1)+0(z−1)=0

3x−9−2y+2=0

3x−2y−7=0

3x−2y=7

This is the required equation of the plane which is parallel to z-axis and passing through the points (3,1,1) and (1,-2,3).