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23.   The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

              (A) Perpendicular          (B) Parallel      (C) intersect y-axis      (D) passes through \left ( 0,0\frac{5}{4} \right )

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)

22.   Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

              (A) 2 units         (B) 4 units           (C) 8 units       (D) \frac{2}{\sqrt{29}}unit

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)   

21.  Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}.

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 

20.  Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

              \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} and \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}

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 =...

19.  Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5 and \overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6.

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

18  Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line \overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) and the plane   \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5.

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.

17.  Find the equation of the plane which contains the line of intersection of the planes \overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0and which is perpendicular to the plane \overrightarrow{r}.(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0

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...

16. If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

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

15.  Find the equation of the plane passing through the line of intersection of the planes \overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1 and \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0   and parallel to x-axis.

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...

14. If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane \overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0 then find the value of p.

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            ...

13. Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

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 

12.  Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

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 .

11.  Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

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 .

10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

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 

9.  Find the shortest distance between lines \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k}) and \overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k}).

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.  

8. Find the equation of the plane passing through (a, b, c) and parallel to the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2.

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...

7.  Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0

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  .

6.  If the lines \frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2} and \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}are perpendicular, find the value of k.

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.  

5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Direction ratios of  AB are  and Direction ratios of CD are  So, it can be noticed that,  Therefore, AB is parallel to CD. Thus, we can easily say the angle between AB and CD which is either .

4. Find the equation of a line parallel to x-axis and passing through the origin.

Equation of a line parallel to the x-axis and passing through the origin  is itself x-axis. So, let A be a point on the x-axis. Therefore, the coordinates of A are given by , where . Now, the direction ratios of OA are  So, the equation of OA is given by,   or   Thus, the equation of the line parallel to the x-axis and passing through origin is 
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