The availability of step-by-step ISC Class 12 Maths OP Malhotra Solutions Chapter 24 The Plane Ex 24(b) can make challenging problems more manageable.
S Chand Class 12 ICSE Maths Solutions Chapter 24 The Plane Ex 24(b)
Question 1.
Find the vector equation of a plane at a distance of 6 units from the origin and has i as the unit vector normal to it.
Answer:
Given p = length of ⊥ origin from plane
= 6 units
\(\hat{n}\) = \(\hat{i}\) = unit vector normal to plane
∴ eqn. of required plane be
\(\vec{r}\) \(\vec{n}\) = p ⇒ \(\vec{r}\) \(\hat{i}\) = 6
Question 2.
Find the vector equation of a plane which is at a distance of
(i) 8 units from the origin and which is normal to the vector \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\).
(ii) 5 units from the origin and which is normal to the vector 2 \(\hat{i}\) + 6 \(\hat{j}\) – 3 \(\hat{k}\).
Answer:
(i) Here p= length of ⊥ from origin to plane =8 units
Here \(\vec{n}\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\) and
\(|\vec{n}|\) = \(|\hat{i}+2 \hat{j}-2 \hat{k}|\)
= \(\sqrt{1^2+2^2+(-2)^2}\) = 3
i.e. \(|\vec{n}|\) < 1
⇒ \(\vec{n}\) is not a unit vector Thus required eqn. of plane be
\(\vec{r}\) \(\vec{n}\) = p
i.e. \(\vec{r}\) \(\frac{(\hat{i}+2 \hat{j}-2 \hat{k})}{3}\) = 8
⇒ \(\vec{r}\) (\(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\)) = 24
(ii) Here p = length of ⊥ from origin to reqd. plane = 5 units
and \(\vec{n}\) = 2 \(\hat{i}\) + 6 \(\hat{j}\) – 3 \(\hat{k}\)
and \(|\vec{n}|\) = |2 \(\hat{i}\) + 6 \(\hat{j}\) – 3 \(\hat{k}\)|
= \(\sqrt{2^2+6^2+(-3)^2}\)
= \(\sqrt{49}\) = 7 < 1
∴ \(\vec{n}\) is not a unit vector.
Thus required eqn. of plane be \(\vec{r}\) \(\hat{n}\) = p
i.e. \(\vec{r}\) \(\frac{(2 \hat{i}+6 \hat{j}-3 \hat{k})}{7}\) = 5
⇒ \(\vec{r}\) (2 \(\hat{i}\) + 6 \(\hat{j}\) – 3 \(\hat{k}\)) = 35
Question 3.
Find the cartesian equation of the following planes
(i) \(\vec{r}\) (\(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) = 1
(ii) \(\vec{r}\) (3 \(\hat{i}\) – 7 \(\hat{j}\) + \(\hat{k}\)) + 3 = 0
(iii) \(\vec{r}\) = (λ + 3µ) \(\hat{i}\) + (2 – µ) \(\hat{j}\) + (µ + 2λ) \(\hat{k}\)
Answer:
(i) Given vector eqn. of plane be
\(\vec{r}\) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) = 1
Since \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point (x, y, z) on plane (1). Thus eqn. (1) becomes:
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) = 1
2 x + 3 y – 4 z = 1 be the required cartesian eqn. a of plane.
(ii) Given vector eqn. of plane be
\(\vec{r}\) (3 \(\hat{i}\) – 7 \(\hat{j}\) + \(\hat{k}\)) + 3 = 0
Since \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point (x, y, z) on plane (1).
Then eqn. (1) becomes;
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (3 \(\hat{i}\) – 7 \(\hat{j}\) + \(\hat{k}\)) + 3 = 0
i.e. 3 x – 7 y + z + 3 = 0
which is the required carterian eqn. of plane.
(iii) \(\vec{r}\) = (λ + 3µ ) \(\hat{i}\) + (2 – µ) \(\hat{j}\) + (µ + 2λ) \(\hat{k}\)
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point on plane (1).
Thus, (x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\))
= (λ + 3µ ) \(\hat{i}\) + (2 – µ) \(\hat{j}\) + (µ + 2λ) \(\hat{k}\)
on comparing the coefficients of \(\hat{i}\), \(\hat{j}\) \(\hat{k}\) on both sides; we have
x = λ + 3 µ
y = 2 – µ
z = µ + 2λ
eqn.(2) + eqn. (3); we have
y + z = 2 + 2λ
⇒ λ = \(\frac{y+z-2}{2}\)
∴ from (1); we have
3 µ = x – \(\frac{y+z-2}{2}\) = \(\frac{2 x-y-z+2}{2}\)
⇒ µ = \(\frac{2 x-y-z+2}{6}\)
∴ from (2);
y = 2 – \(\frac{2 x-y-z+2}{6}\)
⇒ 6 y = 12 – 2 x + y + z – 2
⇒ 2 x + 5 y – z = 10
which is the required cartersian eqn. of plane.
Question 4.
Find the vector equation of the following planes
(i) 5 x – 7 y + 2 z = 3
(ii) x – 2 y + 3 z + 1 = 0
Answer:
(i) Given eqn. of plane be
5 x – 7 y + 2 z = 3
Let x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point on plane (1).
∴ eqn. (1) can be written are :
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (5 \(\hat{i}\) – 7 \(\hat{j}\) + 2 \(\hat{k}\)) = 3
⇒ \(\vec{r}\) (5 \(\hat{i}\) – 7 \(\hat{j}\) + 2 \(\hat{k}\)) = 3
(ii) Given eqn. of plane in cartesian form be
x – 2 y + 3 z + 1 = 0
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point (x, y, z) on plane (1). Then eqn. (1) can be written as:
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (\(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)) + 1 = 0
⇒ \(\vec{r}\) (\(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)) = -1
which is the required vector eqn. of plane.
Question 5.
Find the direction cosines of the perpendicular from the origin to the plane
(i) \(\vec{r}\) (2 \(\hat{i}\) – 3 \(\hat{j}\) – 6 \(\hat{k}\)) + 5 = 0
(ii) \(\vec{r}\) (2 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)) + 2 = 0
Answer:
(i) Given vector eqn. of plane be
\(\vec{r}\) (2 \(\hat{i}\) – 3 \(\hat{j}\) – 6 \(\hat{k}\)) + 5 = 0
⇒ \(\vec{r}\) (2 \(\hat{i}\) – 3 \(\hat{j}\) – 6 \(\hat{k}\)) = -5
The normal form of the plane be
\(\vec{r}\) \(\hat{n}\) = p
where p = + ve.
So eqn. (1) can be written as :
\(\vec{r}\) (-2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\)) = 5
Here \(\vec{n}\) = -2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\) and
\(|\vec{n}|\) = \(\sqrt{4+9+36}\) = 7 < 1
∴ \(\vec{n}\) is not a unit vector. Thus eqn. (2) can be written as
\(\vec{r}\) \(\frac{(-2 \hat{i}+3 \hat{j}+6 \hat{k})}{|\vec{n}|}\) = \(\frac{5}{|\vec{n}|}\)
i.e. \(\vec{r}\) \(\frac{(-2 \hat{i}+3 \hat{j}+6 \hat{k})}{7}\) = \(\frac{5}{7}\)
which is the req. normal form of plane. Then direction cosines of the ⊥ from origin to plane are < \(\frac{-2}{7}\), \(\frac{3}{7}\), \(\frac{6}{7}\) >
(ii) Given vector eqn. of plane be
\(\vec{r}\) (2 \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)) = -2
The normal form of plane be
\(\vec{r}\) \(\hat{n}\) = p
where p be positive.
but here p = -2 < 0
So eqn. (1) can be written as :
\(\vec{r}\) (-2 \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)) = 2
Further \(\vec{n}\) = -2 \(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)
\(|\vec{n}|\) = \(\sqrt{4+4+1}\) = 3 < 1
∴ \(\vec{n}\) is not a unit vector.
So dividing eqn. (2) throughout by \(|\vec{n}|\) i.e. 3 ; we get
\(\vec{r}\) \(\frac{(-2 \hat{i}+2 \hat{j}-\hat{k})}{3}\) = \(\frac{2}{3}\),
Here \(\hat{n}\) = \(-\frac{2}{3}\) \(\hat{i}\) + \(\frac{2}{3}\) \(\hat{j}\) – \(\frac{\hat{k}}{3}\)
Thus direction cosines of the ⊥ from (0,0,0) to given plane are
< \(-\frac{2}{3}\), \(\frac{2}{3}\), – \(\frac{1}{3}\) >
Question 6.
Reduce the equation to normal form and hence find length of perpendicular from the origin to the plane.
(i) \(\vec{r}\) (3 \(\hat{i}\) – 4 \(\hat{j}\) + 12 \(\hat{k}\)) = 5
(ii) \(\vec{r}\) (4 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\)) = 18
Answer:
(i) Given vector eqn. of plane be
\(\vec{r}\) (3 \(\hat{i}\) – 4 \(\hat{j}\) + 12 \(\hat{k}\)) = 5
on comparing with \(\vec{r}\) \(\vec{n}\) = d
i.e. \(\vec{n}\) = 3 \(\hat{i}\) – 4 \(\hat{j}\) + 12 \(\hat{k}\)
and \(|\vec{n}|\) = \(\sqrt{3^2+(-4)^2+12^2}\)
= \(\sqrt{169}\) = 13 < 1
∴ \(\vec{n}\) is not a unit vector and required unit vector
\(\hat{n}\) = \(\frac{\vec{n}}{|\vec{n}|}\) = \(\frac{3 \hat{i}-4 \hat{j}+12 \hat{k}}{13}\)
Thus normal form of plane be
\(\vec{r}\) \(\frac{(3 \hat{i}-4 \hat{j}+12 \hat{k})}{13}\) = \(\frac{5}{13}\)
Thus length of ⊥ from (0,0,0) to given plane = \(\frac{5}{13}\) units
(ii) Given vector eqn. of plane be
\(\vec{r}\)(4 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\)) = 18
On comparing with \(\vec{r}\) \(\vec{n}\) = d
More \(\vec{n}\) = 4 \(\hat{i}\) – 2 \(\hat{j}\) + 4 \(\hat{k}\)
and
\(|\vec{n}|\) = \(\sqrt{4^2+(-2)^2+4^2}\)
= \(\sqrt{36}\) = 6 < 1
∴ \(\vec{n}\) is not a unit vector and reqd. unit vector
\(\vec{n}\) = \(\frac{\vec{n}}{|\vec{n}|}\) = \(\frac{4 \hat{i}-2 \hat{j}+4 \hat{k}}{6}\)
Thus required normal form of given plane be given by
\(\vec{r}\) \(\frac{(4 \hat{i}-2 \hat{j}+4 \hat{k})}{6}\) = \(\frac{18}{6}\) = 3
Thus, reqd. length of ⊥ from origin to given plane = 3
Question 7.
The vector equation of a plane is
\(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) = 9 , where (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) is normal to the plane. Find the length of perpendicular from the origin to the plane.
Answer:
Given vector eqn. of plane be
\(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) = 9
Here \(\vec{n}\) = normal vector to plane (1)
= 2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)
and \(|\vec{n}|\) = \(\sqrt{2^2+(-1)^2+2^2}\) = \(\sqrt{9}\) = 3 < 1
Thus \(\vec{n}\) is not a unit vector.
∴ required unit vector
\(\vec{n}\) = \(\frac{\vec{n}}{|\vec{n}|}\) = \(\frac{2 \hat{i}-\hat{j}+2 \hat{k}}{3}\)
Thus eqn. (1) can be written as :
\(\vec{r}\) \(\frac{(2 \hat{i}-\hat{j}+2 \hat{k})}{3}\) = \(\frac{9}{3}\) = 3
which is of the normal form \(\vec{r}\) \(\vec{n}\) = p
Thus, p = length of ⊥ from origin to given plane =3 units.
Question 8.
Find the vector equation of a plane which is at a distance of 5 units from the origin and has -1,2,2 as the direction ratios of a normal to it.
Answer:
Given p = length of ⊥ from (0,0,0) to required plane = 5 units.
Given direction ratios of normal to plane are < -1, 2, 2 >
∴ direction cosines of normal to plane are
< \(\frac{-1}{\sqrt{1+4+4}}\), \(\frac{2}{\sqrt{1+4+4}}\), \(\frac{2}{\sqrt{1+4+4}}\) >
< \(-\frac{1}{3}\), \(\frac{2}{3}\), \(\frac{2}{3}\) >
Thus \(\hat{n}\) = \(-\frac{\hat{i}}{3}\) + \(\frac{2}{3}\) \(\hat{j}\) + \(\frac{2}{3}\) \(\hat{k}\)
Hence the required normal form of plane be
\(\vec{r}\) \(\hat{n}\) = p \(\vec{r}\) (\(-\frac{1}{3}\) \(\hat{i}\) + \(\frac{2}{3}\) \(\hat{j}\) + \(\frac{2}{3}\) \(\hat{k}\)) = 5
Question 9.
Find a unit normal vector to the plane,
x + 2 y + 3 z – 5 = 0
Answer:
Given cartesian eqn. of plane be
x + 2 y + 3 z – 5 = 0
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point (x, y, z) on plane (1).
∴ eqn. (1) can be written as :
(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) – 5 = 0
i.e. \(\vec{r}\) (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) = 5
which is of the form \(\vec{r}\) \(\vec{n}\) = 1
Here \(\vec{n}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
and \(\vec{n}\) = \(\sqrt{1^2+2^2+3^2}\) = \(\sqrt{14}\) < 1
∴ \(\vec{n}\) is not a unit vector.
∴ required unit vector
= \(\hat{n}\) = \(\frac{\vec{n}}{|\vec{n}|}\) = \(\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{1^2+2^2+3^2}}\)
∴ \(\hat{n}\) = \(\frac{1}{\sqrt{14}}\) \(\hat{i}\) + \(\frac{2}{\sqrt{14}}\) \(\hat{j}\) + \(\frac{3}{\sqrt{14}}\) \(\hat{k}\)
Question 10.
Find the vector equation of a plane passing through a point having position vector 2 i + 3 j – 4 k and perpendicular to the vector 2 i – j + 2 k. Also reduce it to Cartesian form.
Answer:
We know that eqn. of plane passing through the point whose P.V. is \(\vec{a}\) and ⊥ to \(\vec{n}\) is
(\(\vec{r}\) – \(\vec{a}\)) \(\vec{n}\) = 0
Here given \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\) and \(\vec{n}\) = 2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)
∴ eqn. (1) becomes,
\(\vec{r}\) \(\vec{n}\) = \(\vec{a}\) \(\vec{n}\)
⇒ \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= (2 \(\hat{i}\) + 3 \(\hat{j}\) – 4 \(\hat{k}\)) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
⇒ \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\))
= 2(2) + 3(-1) – 4(2)
= -7
which is the required vector eqn. of plane.
Let \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point (x, y, z) on required plane (2).
∴(x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)) (2 \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\)) = -7
⇒ 2 x – y + 2 z + 7 = 0
which is the required cartesian eqn. of plane.
Question 11.
Find the equation of the plane through the points 2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\) and perpendicular to the 3 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\). Determine the perpendicular distance of this plane from the origin.
Answer:
We know that eqn. of plane passing through to the point whose P.V. \(\vec{a}\) and normal to the vector \(\vec{n}\) is given by
\(\vec{r}\) \(\vec{n}\) = \(\vec{a}\) \(\vec{n}\)
Here \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\) and
\(\vec{n}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)
∴ eqn. (1) becomes ;
\(\vec{r}\) (3 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\))
= (2 \(\hat{i}\) + 3 \(\hat{j}\) – \(\hat{k}\))(3 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\))
= 2(3) + 3(-2) – 1(-2) = 2
Which is the required vector eqn. of plane p = length of ⊥ from (0, 0, 0) to plane (2)
= \(\frac{|\vec{a} \vec{n}|}{|\vec{n}|}\) = \(\frac{2}{\sqrt{3^2+(-2)^2+(-2)^2}}\)
= \(\frac{2}{\sqrt{17}}\) units
Question 12.
Find the equation of the plane through the points 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\) and perpendicular to the vector 6 \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\). Put the above equation in normal form.
Answer:
We know that eqn. of plane passing through the point whose P.V. is \(\vec{a}\) and normal to \(\vec{n}\) is given by
(\(\vec{r}\) – \(\vec{a}\)) \(\vec{n}\) = 0
⇒ \(\vec{r}\) \(\vec{n}\) = \(\vec{a}\) \(\vec{n}\)
Here \(\vec{a}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)
and \(\vec{n}\) = 6 \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\)
∴ eqn. (1) becomes;
\(\vec{r}\) (6 \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\))
= (2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)) (6 \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\))
= 2(6) + 3(4) + 4(3) = 36
Which is of the form \(\vec{r}\) \(\vec{n}\) = d
Here \(\vec{n}\) = 6 \(\hat{i}\) + 4 \(\hat{j}\) + 3 \(\hat{k}\)
i.e. \(|\vec{n}|\) = \(\sqrt{36+16+9}\) = \(\sqrt{61}\) < 1
∴ \(\vec{n}\) is not a unit vector.
Thus the required normal form of plane be
\(\vec{r}\) \(\frac{\vec{n}}{|\vec{n}|}\) = \(\frac{d}{|\vec{n}|}\)
i.e. \(\vec{r}\) \(\frac{(6 \hat{i}+4 \hat{j} + 3 \hat{k})}{\sqrt{61}}\) = \(\frac{36}{\sqrt{61}}\)
Question 13.
Find the vector equations of the coordinate planes.
Answer:
(i) eqn. of XOY plane be z = 0
i.e. \(\vec{r}\) \(\vec{k}\) = 0
Where \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point (x, y, z) on XOY plane.
(ii) eqn. of YOZ plane be x = 0 i.e. \(\vec{r}\) \(\hat{i}\) = 0
Where \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)
(iii) eqn. of ZOX plane be y = 0 i.e. \(\vec{r}\) \(\hat{j}\) = 0
Where \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\)
Question 14.
Show that the normals to the following pairs of plants are perpendicular to each other.
(i) x – y – z – 2 = 0 and 3 x + 2 y + z + 4 = 0
(ii) \(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) = 5 and
\(\vec{r}\) (2 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) = 5
Answer:
(i) Given eqns. of planes are
x – y – z – 2 = 0
and 3 x + 2 y + z + 4 = 0
Vector form of planes (1) and (2) are :
\(\vec{r}\) (\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) = 2
and
\(\vec{r}\) (3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) + 4 = 0
Here
\(\vec{n}_1\) = \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)
\(\vec{n}_2\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
\(\vec{n}_1\) \(\vec{n}_2\) = (\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) (3 \(\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\))
= 1(3) – 1(2) – 1(1) = 0
Thus \(\vec{n}_1\) ⊥ \(\vec{n}_2\)
(ii) Given eqn. of planes are
\(\vec{r}\) (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) = 5
and \(\vec{r}\) (2 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)) = 5
Let \(\vec{n}_1\) and \(\vec{n}_2\) be the vectors normal to given planes (1) and (2).
i.e. \(\vec{n}_1\) = 2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)
and \(\vec{n}_2\) = 2 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)
Here,
\(\vec{n}_1\) \(\vec{n}_2\) = (2 \(\hat{i}\) – \(\hat{j}\) + 3 \(\hat{k}\)) (2 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\))
= 2(2) – 1(-2) + 3(-2)
= 6 – 6 = 0
Thus \(\vec{n}_1\) ⊥ \(\vec{n}_2\)
Question 15.
Show that the normal vector to the plane 2 x + 2 y + 2 z = 3 is equally inclined with the coordinate axes.
Answer:
Given eqn. of plane be
2 x + 2 y + 2 z = 3
So its vector form be
\(\vec{r}\) (2 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)) = 3
Where \(\vec{r}\) = x \(\hat{i}\) + y \(\hat{j}\) + z \(\hat{k}\) be the P.V. of any point on plane (1).
Eqn. (2) is the form
\(\vec{r}\) \(\vec{n}\) = d
i.e. \(\vec{n}\) = vector normal to plane (2)
= 2 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)
Let α, β, γ be the angle made by \(\vec{n}\) with x axis, y-axis and z-axis respectively.
∴ cosα = \(\frac{\vec{n} \hat{i}}{|\vec{n}||\hat{i}|}\) = \(\frac{(2 \hat{i}+2 \hat{j}+2 \hat{k}) \hat{i}}{\sqrt{4+4+4} 1}\)
= \(\frac{2}{\sqrt{12}}\) = \(\frac{2}{2 \sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)
and
cos β = \(\frac{\vec{n} \cdot \hat{j}}{|\vec{n}||\hat{j}|}\) = \(\frac{(2 \hat{i}+2 \hat{j}+2 \hat{k}) \cdot \hat{j}}{\sqrt{4+4+4} \cdot 1}\)
= \(\frac{2}{2 \sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)
Also, cosα = \(\frac{\vec{n} \cdot \hat{k}}{|\vec{n}||\hat{k}|}\) = \(\frac{(2 \hat{i}+2 \hat{j}+2 \hat{k}) \cdot \hat{k}}{\sqrt{4+4+4} \cdot 1}\)
= \(\frac{2}{\sqrt{12}}\) = \(\frac{2}{2 \sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)
∴ cosα = cosβ = cosγ = \(\frac{1}{\sqrt{3}}\)
⇒ α = β = γ
Hence, the normal vector to given plane is equally inclined to coordinate axes.