15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T682,

The equation of the given first plane is

$$x+y+z-1=0$$ .... (i)

where (1, 1, 1) are direction ratio of normal to the plane.

The equation of second plane is

$$2x+3y-z+4=0$$ ..... (ii)

where (2, 3, $$-$$1) are direction ratio of normal to the second plane.

We know that, the equation of any plane passes through the intersection of two given planes

$$ax+by+cz+d=0$$ and $$a_1x+b_1y+c_1z+d_1=0$$ is

$$ax+by+cz+d+\lambda(a_1x+b_1y+c_1z+d_1)=0$$ ... (iii)

where $$\lambda$$ is arbitrary constant.

Here, $$a=1,b=1,c=1,d=-1$$

$$a_1=2,b_1=3,c_1=-1,d_1=4$$

Putting these values in Eq. (iii), we get

$$x+y+z-1+\lambda(2x+3y-z+4)=0$$

$$x(1+2\lambda)+y(1+3\lambda)+z(1-\lambda)-1+4\lambda=0$$ ..... (iv)

It is given that the plane (iv) is parallel to the X-axis, then the plane (iv) is perpendicular to the YZ plane. Therefore, the normal of the plane (iv) and the YZ-plane are also perpendicular to each other.

Equation of YZ plane is $$x = 0$$ i.e.

$$1\,.\,x + 0\,.\,y + 0\,.\,z = 0$$

$$ \Rightarrow 1(1 + 2\lambda ) + 0(1 + 3\lambda ) + 0(1 - \lambda ) = 0$$

$$1 + 2\lambda = 0 \Rightarrow \lambda = - {1 \over 2}$$

Put the value of $$\lambda = - {1 \over 2}$$ in Eq. (iv), we get

$$x\left( {1 - {2 \over 2}} \right) + y\left( {1 - {3 \over 2}} \right) + z\left( {1 + {1 \over 2}} \right) - 1 + 4\left( { - {1 \over 2}} \right) = 0$$

$$x\,.\,0 + y\left( { - {1 \over 2}} \right) + {{3z} \over 2}

Three Dimensional Geometry

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