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Consider the lines
Q. The unit vector perpendicular to both L_{1} and L_{2} is
Vector in the direction of
Vector in the direction of
∴ Vector perpendicular to both L_{1} and L_{2}
Consider the lines
Q. The shortest distance between L_{1} and L_{2} is
The shortest distance between L_{1} and L_{2} is
Consider the lines
Q. The distance of the point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to both the lines L_{1} and L_{2} is
The plane passing through (–1, –2, –1) and having normal along
– 1(x + 1) – 7(y + 2) + 5(z + 1) = 0
or x + 7y  5z +10 = 0
∴ Distance of point (1, 1, 1) from the above plane is
Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5.
STATEMENT1 : The parametric equations of the line of intersection of the given planes are x = 3 + 14t, y = 1 + 2t, z = 15t. because
STATEMENT2 : The vector is parallel to the line of intersection of given planes.
The line of intersection of given plane is 3x 6y  2z 15 = 0 = 2x + y  2z5
For z = 0 , we obtain x = 3 and y =1
∴Line passes through (3, –1, 0).
Let a, b, c be the d’rs of line of intersection, then 3a 6b 2c = 0 and 2a + b  2c = 0
Solving the above equations using cross product method, we get a :b :c = 14 : 2 : 15
∴ Eqn. of line is
whose parametric form is x = 3 + 14t , y = 1 + 2t, z= 15t
∴ StatementI is false (value of y is not matching).
Since dr’s of line intersection of given planes are 14, 2,15
∴ is parallel to this line.
∴ Statement 2 is true.
Let the vectors epresent the sides of a regular hexagon.
STATEMENT1 :
STATEMENT2 :
Statement1 is true.
∴ Statement2 is false.
Consider three planes
P_{1} : x – y + z = 1
P_{2} : x + y – z = 1
P_{3} : x – 3y + 3z = 2
Let L_{1}, L_{2}, L_{3} be the lines of intersection of the planes P_{2} and P_{3}, P_{3} and P_{1}, P_{1} and P_{2}, respectively.
STATEMENT  1Z : At least two of the lines L_{1}, L_{2} and L_{3} are nonparallel and
STATEMENT  2 : The three planes doe not have a common point.
The given planes are
P_{1} : x – y + z = 1 ...(1)
P_{2} : x + y – z = –1 ...(2)
P_{3} : x – 3y + 3z = 2 ...(3)
Line L_{1} is intersection of planes P_{2} and P_{3}.
∴ L_{1} is parallel to the vector
Line L_{2} is intersection of P_{3} and P_{1}
∴ L_{2} is parallel to the vector
Line L_{3} is intersection of P_{1} and P_{2}
∴ L_{3} is parallel to the vector
Clearly lines L_{1}, L_{2} and L_{3} are parallel to each other.
∴ Statement–1 is False
Also family of planes passing through the intersection of P_{1} and P_{2}_{ }is P_{1} + λP_{2} = 0 .If plane P_{3} is represented by P_{1} + λP_{2} = 0 for some value of l then the three planes pass through the same point.
Here P_{1} + λP_{2} = 0
This will be identical to P_{3} if
...(1)
and taking
∴There is no value of λ which satisfies eq (1).
∴ The three planes do not have a common point.
⇒ Statement 2 is true.
∴ (d) is the correct option.
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