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G Gautam harsolia
Given equations of planes are and  Now, from equation (i) and (ii) it is clear that given planes are parallel to each other  Therefore, the correct answer is (B)

G Gautam harsolia
Given equations are  and  Now, it is clear from  equation (i) and (ii) that given planes are parallel We know that the distance between two parallel planes   is given by  Put the values in this equation we will get, Therefore, the correct answer is (D)

G Gautam harsolia
The equation of plane having a, b and c intercepts with x, y  and z-axis respectively is given by The distance p of the plane from the origin is given by  Hence proved

P Pankaj Sanodiya
Given  Two straight lines in 3D whose direction cosines (3,-16,7) and (3,8,-5) Now the two vectors which are parallel to the two lines are  and  As we know, a vector perpendicular to both  vectors  and  is , so A vector parallel to this vector is  Now as we know the vector equation of the line which passes through point p and parallel to vector d is Here in our question, give point p =...

P Pankaj Sanodiya
Given  A point through which line passes two plane    And   it can be seen that normals of the planes are since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes. So, a vector perpendicular to both these normal vector is Now a line which passes through  and parallels to  is  So the required line is

P Pankaj Sanodiya
Given,  Equation of a line : Equation of the plane Let's first find out the point of intersection of line and plane. putting the value of  into the equation of a plane from the equation from line Now, from the equation, any  point p  in line is  So the point of intersection is  SO, Now, The distance between the points (-1,-5,-10) and (2,-1,2) is  Hence the required distance is 13.

D Divya Prakash Singh
The equation of the plane passing through the line of intersection of the given plane in          ,,,,,,,,,,,,,(1) The plane in equation (1) is perpendicular to the plane,  Therefore  Substituting  in equation (1), we obtain                      .......................(4) So, this is the vector equation of the required plane. The Cartesian equation of this plane can be obtained by...

D Divya Prakash Singh
We have the coordinates of the points   and   respectively. Therefore, the direction ratios of OP are  And we know that the equation of the plane passing through the point  is  where a,b,c are the direction ratios of normal. Here, the direction ratios of normal are  and  and the point P is . Thus, the equation of the required plane is

D Divya Prakash Singh
So, the given planes are:   and   The equation of any plane passing through the line of intersection of these planes is             ..............(1) Its direction ratios are   and   = 0  The required plane is parallel to the x-axis. Therefore, its normal is perpendicular to the x-axis. The direction ratios of the x-axis are 1,0, and 0. Substituting  in equation (1), we obtain So, the...

D Divya Prakash Singh
Given that the points  and  are equidistant from the plane So we can write the position vector through the point  is  Similarly, the position vector through the point  is The equation of the given plane is  and We know that the perpendicular distance between a point whose position vector is   and the plane,   and  Therefore, the distance between the point  and the given plane is            ...

P Pankaj Sanodiya
Given  two planes x + 2y + 3z = 5 and 3x + 3y + z = 0. the normal vectors  of these plane are  Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be : Now, as we know  the equation of a plane in vector form is : Now Since this plane passes through the point (-1,3,2) Hence the equation of the plane is

D Divya Prakash Singh
We know that the equation of the line that passes through the points and  is given by the relation; and the line passing through the points, .    And any point on the line is of the form. This point lies on the plane,  or  . Hence, the coordinates of the required point are   or .

D Divya Prakash Singh
We know that the equation of the line that passes through the points and  is given by the relation; and the line passing through the points,     And any point on the line is of the form . So, the equation of ZX plane is  Since the line passes through YZ- plane, we have then, or     and   So, therefore the required point is .

D Divya Prakash Singh
We know that the equation of the line that passes through the points and  is given by the relation; and the line passing through the points,     And any point on the line is of the form . So, the equation of the YZ plane is  Since the line passes through YZ- plane, we have then, or     and   So, therefore the required point is

D Divya Prakash Singh
Given lines are;    and So, we can find the shortest distance between two lines  and  by the formula,                                                ...........................(1) Now, we have from the comparisons of the given equations of lines.                     So,  and  Now, substituting all values in equation (3) we get, Hence the shortest distance between the two given lines is 9 units.

D Divya Prakash Singh
Given that the plane is passing through  and is parallel to the plane  So, we have The position vector of the point  is,  and any plane which is parallel to the plane,  is of the form, .                      .......................(1) Therefore the equation we get, Or,  So, now substituting the value of  in equation (1), we get             .................(2) So, this is the required equation...

D Divya Prakash Singh
Given that the plane is passing through the point  so, the position vector of the point A is   and perpendicular to the plane  whose direction ratios are   and the normal vector is  So, the equation of a line passing through a point and perpendicular to the given plane is given by, , where  .

D Divya Prakash Singh
Given both lines are perpendicular so we have the relation;  For the two lines whose direction ratios are known,   We have the direction ratios of the lines,   and    are   and  respectively. Therefore applying the formula,   or    For,  the lines are perpendicular.