Ans. We know that direction ratios of the line joining the origin (0, 0, 0) to the point are 

 = 2 – 0, 1 – 0, 1 – 0 = 2, 1, 1 = 

Similarly, direction ratios of the line joining the points  and  are

 =  =  = 

For these two lines,

=  = 2 – 2 + 0 = 0

Therefore, the two given lines are perpendicular to each other.

Ans.  and  are direction cosines of two mutually perpendicular of two given lines L₁ and L₂. (say) 

Let  and  be the unit vectors along these lines L₁ and L₂.

  and 

Let L be the line perpendicular to both the lines L₁ and L₂ and let  be a unit vector along line L perpendicular both lines L₁ and L₂.

 Cross-product of two vectors = 

[ L1  L2 (given,  angle between them is ]

 

Since, is a unit vector, therefore its components are its direction cosines.

Thus, direction cosines of  are 

direction cosines of line L are 

Ans. Direction ratios of one line are  

A vector along this line is 

Direction ratios of second line are 

A vector along second line is 

Let  be the angle between the two lines, then

 =

=  = 0 = 

Ans. We know that a unit vector along axis is  

 Direction cosines of axis are coefficients of  in the unit vector

i.e., 1, 0, 0 = 

Equation of the required line passing through the origin (0, 0, 0) and parallel to axis is   

Vector equation of the required line is 

 [ and  ]

Ans. Given: Points A B C and D 

 Direction ratios of line AB are 

A vector along the line AB is 

Similarly, direction ratios of line CD are 

A vector along the line AB is 

Let  be the angle between the two lines, then

 = =  =  =  = 1

= 

Therefore, lines AB and CD are parallel.

Ans. Given: Equation of one line is  

Direction ratios of this line are its denominators, i.e.,  = 

A vector along this line is 

Again, equation of second line is 

Direction ratios of this line are its denominators, i.e.,  = 

A vector along this line is 

Since these given lines are perpendicular.

  

 

Ans. The required line passes through the point P (1, 2, 3). 

 Position vector  (say) of point P is (1, 2, 3)

 

Equation of the given plane is 

Comparing with  

Since, the required line is perpendicular to the given plane, therefore, vector  along the required line is 

 Equation of the required line is 

 

Ans. Equation of any plane parallel to the plane  is …..(i) 

Plane (i) passes through 

Putting  in eq. (i), we get

  

Putting the value of  in eq. (i), to get the required plane is

Ans. Given: Vector equation of one line is  

Comparing with  we get

 and 

Again given: Vector equation of another line is 

Comparing with  we get

 and 

We know that length of shortest distance between two (skew) lines is  ..(i)

Now =  = 

Again

Expanding along first row,

= 

 

= 

And = 

Putting these values in eq. (i), length of shortest distance = 

Ans. Given: A line through the points A (5, 1, 6) and B (3, 4, 1) 

 Direction ratios of this line AB are 

 3 – 5, 4 – 1, 1 – 6 

Equation of the line AB is 

 ……….(i)

Now we have to find the coordinates of the point where this line AB crosses the YZ-plane

i.e.,  ……….(ii)

Putting  in eq. (i), we get

  and 

 and   and 

 and 

Thus, required point is P

Ans. Given: A line through the points A (5, 1, 6) and B (3, 4, 1) 

 Direction ratios of this line AB are 

3 – 5, 4 – 1, 1 – 6 

Equation of the line AB is 

 ……….(i)

Now we have to find the coordinates of the point where this line AB crosses the ZX-plane

i.e.,  ……….(ii)

Putting  in eq. (i), we get

  and 

 and   and 

 and 

Thus, required point is P

Ans. Direction ratios of the line joining the points A and B are 

 

 Equation of the line AB are  ………(i)

Equation of the plane is  ………(ii)

Now to find the point where line (i) crosses plane (ii),

From eq. (i) (say)

  

 ……….(iii)

Putting the values of  in eq. (ii), we get

  

Putting  in eq. (iii), point of intersection of line (i) and plane (ii) is

Thus, required point of intersection is 

Ans. Since equation of any plane through the point  is a(x−x1)+b(y−y1)+c(z−z1)=0a(x−x1)+b(y−y1)+c(z−z1)=0 

  ……….(i)

  

This required plane is perpendicular to the plane  

Product of coefficients  ……….(ii)

Again the required plane is perpendicular to the plane 

Product of coefficients  ……….(iii)

Solving eq. (ii) and (iii), we get

Putting these values of  in eq. (i), we get

Ans. Equation of the given plane is  

[  = Position vector of any point  on the plane ]

 ……….(i)

Also, the point  and  are equidistant from plane (i)

(Perpendicular) distance of point  from plane (i)

= Distance of point  from plane (i)

 [ If  then]

Taking positive sign,

Taking negative sign, 

Hence, the values of  are 1 or 

Ans. Equation of one plane is  

 ……….(i)

Equation of the second plane is ……….(ii)

Since, equation of any plane passing through the line intersection of these two planes is

L.H.S. of I +  (L.H.S. of II) = 0

r→.[iˆ+jˆ+kˆ+λ.(2iˆ+3jˆ−kˆ)]−1+4λ=0r→.[i^+j^+k^+λ.(2i^+3j^−k^)]−1+4λ=0

 ……….(i)

Comparing  we have

Now required plane (i) is parallel to axis ( a vector  along axis is  = ) 

  

Putting  in eq. (i), the equation of required plane,

Ans. Given: Origin O (0, 0, 0) and point P 

To find: Equation of the plane passing through P = 

 Direction ratios of normal OP to the plane are 

 

Equation of the required plane is 

Ans. Equation of any plane passing through (or containing) the line of intersection of the planes  and r→(2iˆ+jˆ−kˆ)+5=0r→(2i^+j^−k^)+5=0 is L.H.S. of I +  (L.H.S. of II) = 0 

 ……….(i)

Comparing with  we have,

Now plane (i) is perpendicular to the given plane 

 r→(5iˆ+3jˆ−6kˆ)=−8r→(5i^+3j^−6k^)=−8

Comparing with  we have,

For perpendicular planes

Putting  in eq. (i), equation of required plane is

Ans. Given: A point P (say)  

and equation of the line  ……….(i)

equation of the plane is 

Putting the value of  from eq. (i) in eq. (ii),

Putting  in eq. (i),

Therefore, Point of intersection is  = 

 Distance of the given point P from the point of intersection is

= 

=  units

Ans. The required line passes through the point A (1, 2, 3) =  

 = Position vector of point A = 

Let  be any vector along the required line.

Vector equation of required line is 

  ……….(i)

Since required line is parallel to the plane 

 and 

Comparing with  we have,

And Comparing with  we have,

Since  is perpendicular to both  and 

 = 

Expanding along first row,

 = 

Putting this value of  in eq. (i), vector equation of required line,

Ans. Given: A point on the required line is A 

 Position vector of point A is 

Also given equations of two lines

 and 

Direction ratios of given two lines are  and 

Now = 

Expanding along first row,

=  = 

 

Equation of the required line is 

Again replacing  by 

Ans. We know that equation of plane making intercepts  (on the axes) is  

Given: Perpendicular distance of the origin (0, 0, 0) from plane = 

  = 

 

Squaring both sides,

Ans. Equation of one plane is   

Equation of second plane is  

Here

Since,  therefore, the given two lines are parallel.

We know that the distance of the parallel lines = 

= 

= 

Therefore, option (D) is correct.

Ans. Equations of the given planes are   

and  

For perpendicular =  = 

  

 Planes are not perpendicular.

For parallel

 given planes are parallel.

Therefore, option (B) is correct.