Chapter 6-Plane Exercise 6.1 ~ Mathematics Answer

LeGeNd.

Exercise-6.1

1.Please copy the question from book.

Soln:

Given plane is;

Or, 2x + 3y + 4z = 12

Or,

\frac{x}{6} + \frac{y}{4} + \frac{z}{3}

= 1.

Comparing with

Or,

\frac{x}{a} + \frac{y}{b} + \frac{z}{c}

= 1.

x - intercept, a = 6 , y – intercept, b = 4 and z – intercept, c = 3.

So, required intercepts are 6,4,3.

Soln:

Given plane is;

Or, 4x + 8y + 8z = 9.

Dividing by

\sqrt{4^{2} + 8^{2} + 8^{2}}

=12

We have,

Or,

\frac{x}{3} + \frac{2 y}{3} + \frac{2 z}{3}

\frac{3}{4}

Which is the required normal form. The d.c.’s of the normal

\frac{1}{3}, \frac{2}{3}, \frac{2}{3}

and the length of perpendicular from origin to the plane is, p =

\frac{3}{4}

2.Please copy the question from book.

Soln:

Given, x – intercept, a = 2, y – intercept, b = 3 and z – intercept , c = 4.

So, the required plane is

\frac{x}{a} + \frac{y}{b} + \frac{z}{c}

= 1.

So,

\frac{x}{2} + \frac{y}{3} + \frac{z}{4}

= 1.

So, 6x + 4y + 3x = 12

Soln:

Let the equal intercept be a.

So, the required plane is

\frac{x}{a} + \frac{y}{b} + \frac{z}{c}

= 1.

So, x + y + z = a …(i)

Since plane (i) passes through (2,3,4),

So, 2 + 3 + 4 = a

So, a = 9.

Thus, from (i), the required plane is, x + y + z = 9.

3.Please copy the question from book.

Soln:
The given points are A(1,1,0), B = (-2,2,-1) and C(1,2,1).

Now, any plane through A(1,1,0) is:

a(x – 1) + b(y – 1) + c(z – 0) = 0 …….(i)

The plant (i) also passes through B and C. so,

Or, a(- 2 – 1) + b(2 – 1) + c( - 1 – 0) = 0

So, - 3a + b – c = 0 …..(ii)

And, a(1 – 1) + b(2 – 1) + c(1 – 0) = 0.

So, a.0 + b + c = 0…… (iii)

From (ii) and (iii) by cross multiplication, we have,

Or,

\frac{a}{1 + 1} = \frac{b}{0 + 3} = \frac{c}{- 3 - 0}

Or,

\frac{a}{2} = \frac{b}{3} = \frac{c}{- 3}

…….(iv)

Now from (i) and (ii), we have,

Or, 2(x – 1) + 3(y – 1) – 3(z – 0) = 0

So, 2x + 3y – 3z = 5.

Soln:

Given points are A(0,-1,-1), B(4,5,1), C(3,9,4) and D(-4,4,4)

To show: A,B,C and D are coplanar.

Now, any plane through A(0,-1,-1) is:

Or, a(x – 0) + b(y + 1) + c(z + 1) = 0 …(i)

The plane (i) passes through B and C, so

Or, a(4 – 0) + b(5 + 1) + c(1 + 1) = 0

So, 4a + 6b + 2c = 0 …..(ii)

And, a(3 – 0) + b (9 + 1) + c(4 + 1) = 0 ……. (ii)

So, 3a + 10b + 5c = 0 …(iii).

From (ii) and (iii), we have,

Or,

\frac{a}{30 - 20} = \frac{b}{6 - 20} = \frac{c}{40 - 18}

So,

\frac{a}{10} = \frac{b}{- 14} = \frac{c}{22}

So,

\frac{a}{5} = \frac{b}{- 7} = \frac{c}{11}

..... …(iv)

Eliminating a,b,c from (i) and (iv), we have,

Or, 5(x – 0) – 7(y + 1) + 11(z + 1) = 0

So, 5x – 7y + 11z + 4 = 0 ….(v)

Is the plane through A,B and C.

Now, putting D(-4,4,4) in (v), we have,

Or, 5 * (-4) – 7 * 4 + 11 * 4 + 4 = - 20 – 28 + 44 + 4 = 0

Hence, the given points are coplanar and they lie on the plane,

5x – 7y + 11z + 4 = 0.

4.Please copy the question from book.

Soln:

Any plane through (3,-4,5) is:

Or, a(x – 3) + b(y + 4) + c(z – 5) = 0

The plane (i) is parallel to the plane,

Or, 3x – 4y + 5x = 7.

So, the normal’s plane (i) and (ii) are parallel.

So,

\frac{a}{3} = \frac{b}{- 4} = \frac{c}{5}

….(iii)

Now, from (i) and (iii) we have,

Or, 3(x – 3) – 4(y + 4) + 5(z – 5) = 0

So, 3x – 4y + 5z – 50 = 0.

Soln:

Given plane is ax + by + cz = 0.

Any plane to the plane (i) is ax + by + cz + k = 0 …(ii)

Where, k is to be determined,

The plane (ii) passes through (α,β,γ),so , aα + bβ + cγ+ k = 0.

So, k = - (aα + bβ + cγ)

Thus, from (ii) the required plane is, ax + by + cz = aα + bβ + cγ.

5.Please copy the question from book.

Soln:

The given plane are, x + 2y + z + 7 = 0 and 2x + y – z + 13 = 0.

Let θ be the angle between the planes,

So, cosθ =

\frac{a_{1} . a_{2} + b_{1} . b_{2} + c_{1} . c_{2}}{\sqrt{(a_{1}^{2} + b_{1}^{2} + c_{1}^{2}) (a_{2}^{2} + b_{2}^{2} + c_{2}^{2})} .}

\frac{1.2 + 2.1 - 1.1}{\sqrt{(1^{2} + 2^{2} + 1^{2}) (2^{2} + 1^{2} + {(- 1)}^{2})}}

So, θ = 60° =

\frac{π}{3}

\frac{π}{3}

The given planes are, x + 2y + 3z = 6 and 3x – 3y + z = 1.

Let θ be the angle between the planes.

So, cosθ =

\frac{1.3 + 2. (- 3) + 3.1}{\sqrt{(1 + 4 + 9) (9 + 9 + 1)}}

= 0

So, θ =

\frac{π}{2} .

\frac{π}{2} .

6.Please copy the question from book.

Soln:

The planes are:

2x + 3y – 4z = 3 ….(i)

10x + 15y – 20z = 12…(ii)

And 3x + 2y + 3z = 5 …(iii)

Now planes (i) and (ii) will be parallel if the direction ratios of the normal’s to the plane (i) and (ii) are proportional.

i.e. if

\frac{2}{10}

\frac{3}{15}

\frac{- 4}{- 20}

i.e.

\frac{1}{5} = \frac{1}{5} = \frac{1}{5} .

Hence, the plane (i) and (ii) are parallel.

Again, the plane (i) and (iii) will be perpendicular,

If a1.a2 + b1b2 + c1c2 = 0

i.e. if 2.3 + 3.2 + (-4).3 = 0

i.e. if 0 = 0.

So, a1.a2 + b1.b2 + c1.c2 = 0

So, the planes (i) and (iii) are perpendicular.

7.Please copy the question from book.

Soln:

Any plane through (1,2,3) is:

Or, a(x – 1) + b(y – 2) + c(z – 3) = 0 …(i)

Since the plane (i) is normal to the planes

x – y – z = 5 …(ii)

And 2x – 5y – 3z = 7 …(iii)

So, a.1 + b(-1) + c(-1) = 0 [from(i) and (ii)]

So, a – b – c = 0 ….(iv)

And 2a – 5b – 3c = 0 ….() [from (i) and (iii)]

So, from (iv) and (v) we have,

\frac{a}{3 - 5} = \frac{b}{- 2 + 3}

\frac{c}{- 5 + 2}

So,

\frac{a}{- 2} = \frac{b}{1} = \frac{c}{- 3}

…(vi)

From (i) and (vi),

We have, - 2(x – 1) + 1(y – 2) – 3(z – 3) = 0

Or, - 2x + 2 + y – 2 – 3z + 9 = 0

So, 2x – y + 3z – 9 = 0.

b.Please copy the question from book.

Soln:

Any plane through (2,1,4) is:

Or, a(x – 2) + b(y – 1) + c(z – 4) = 0….(i)

Since, the plane (i) is perpendicular to the planes.

9x – 7y + 6z + 48 = 0 ….(ii)

And x + y + z = 0 ….(iii)

So, we have, 9a – 7b + 6c = 0

And a + b + c = 0.

So,

\frac{a}{- 7 - 6} = \frac{b}{6 - 9} = \frac{c}{9 + 7}

So,

\frac{a}{- 13} == \frac{b}{- 3} = \frac{c}{16}

…(iv)

Thus, from (i) and (iv) we have, - 13(x – 2) – 3(y – 1) + 16(z – 4) = 0.

Or, - 13x + 26 – 3y + 3 + 16z – 64 = 0.

So, 13x + 3y – 16z + 35 = 0.

8.Please copy the question from book.

Soln:

Any plane through P(a,b,c) is:

Or, l(x – a) + m(y – b) + n(z – c) = 0 …(i)

Since the plane (i) is perpendicular to the line OP joining O(0,0,0) and P(a,b,c) so the line normal to the plane (i) is parallel to the line OP, so we have,

Or,

\frac{l}{a - 0} = \frac{m}{b - 0} = \frac{n}{c - 0}

So,

\frac{l}{a} = \frac{m}{b} = \frac{n}{c}

….(ii)

Thus, from (i) and (ii)

We have, a(x – a) + b(y – b) + c(z – c) = 0.

So, ax + by + cz = a2 + b2 + c2, is the required plane.

Soln:

Any plane through (2,-3,1) is

a (x – 2) + b(y + 3) + c(z – 1) = 0 ….(i)

Since the plane (i) is perpendicular to the line joining the points (3,4,-1) and B(2,-1,5) so the line normal to the plane (i) is parallel to the line AB, so we have,

Or,

\frac{a}{3 - 2} = \frac{b}{4 + 1} = \frac{c}{- 1 - 5}

[3 – 4,4 + 1, - 1 – 5 are dr’s of AB]

So,

\frac{a}{1} = \frac{b}{5} = \frac{c}{- 6}

Thus, from (i) and (ii) we have,

1(x – 2) + 5(y + 3) – 6(z – 1) = 0

So, x + 5y – 6z + 19 = 0.

9.Please copy the question from book.

Soln:

Any plane (2,2,1) is a(x – 2) + b(y – 2) + c(z – 1) = 0 ….(1)

The plane (1) passes through (9,3,6), so 7a + b + 5c = 0 ….(2)

The plane (1) is normal to the plane 2x + 6y + 6z + 9 , so

2a + 6b + 6c = 0 …(3) [a1.a2 + b1.b2 + c1.c2 = 0]

From (2) and (3) we have,

\frac{a}{6 - 30} = \frac{b}{10 - 42} = \frac{c}{42 - 2}

Or,

\frac{a}{3} = \frac{b}{4} = \frac{c}{- 5}

. …(4)

Eliminating a,b,c from (1) and (4), so,

So, 3(x – 2) + 4(y – 2) – 5(z – 1) = 0.

So, 3x + 4y – 5z – 9 = 0.

Soln:

Any plane (-1,1,-1) is a(x + 1) + b(y – 1) + c(z + 1) = 0 ….(1)

The plane (1) passes through (6,2,1), so 7a + b + 2c = 0 ….(2)

The plane (1) is normal to the plane 2x + y + z = 5 , so

2a + b + c = 0 …(3)

From (2) and (3) we have,

\frac{a}{1 - 2} = \frac{b}{4 - 7} = \frac{c}{7 - 2}

\frac{a}{1} = \frac{b}{3} = \frac{c}{- 5}

Eliminating a,b,c from (1) and (4), so,

So, 1(x + 1) + 3(y – 1) – 5(z + 1) = 0.

So, x + 3y – 5z = 7.

10.Please copy the question from book.

Soln:

Any plane through the planes x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 is

X + y + z – 6 + k(2x + 3y + 4z + 5) = 0 ….(i)

Where, k is to be determined.

So, (1 + 2k)x + (1 + 3k)y + (1 + 4k)z + (-6 + 5k) = 0 ….(2)

The plane (2) is perpendicular to the plane 4x + 5y – 3z = 8, so,

Or, 4(1 + 2k) + 5(1 + 3k) – 3(1 + 4k) = 0

Or, 6 + 11k = 0

So, k =

- \frac{6}{11}

Thus, from (1), the required plane is,

Or, 11(x + y + z – 6) – 6(2x + 3y + 4z + 5) = 0

So, x + 7y + 13z + 96 = 0.

11.Please copy the question from book.

Soln

(i)

Given plane is 6x + 6y + 3z – 11 = 0 …(1)

So, the perpendicular distance from (2,5,7) to (1) is:

P =

\frac{| 6.2 + 6.5 + 3.7 - 11 |}{\sqrt{6^{2} + 6^{2} + 3^{2}}}

\frac{52}{9}

(ii) Required length, p =

\frac{| 3.0 - 2.0 + 6.0 - 17 |}{\sqrt{9 + 4 + 36}}

\frac{17}{7}

units.

12.

Soln:

(i)

The given planes are, x – y + 2z – 4 = 0 ….(1)

And, 2x – 2y + 4z + 5 = 0
So, x – y + 2z +

\frac{5}{2}

= 0 ….(2)

Since, the planes (1) and (2) are parallel, so the distance between them ;s:

d =

\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2} + c^{2}}}

\frac{| - 4 - \frac{5}{2} |}{\sqrt{1^{2} + {(- 1)}^{2} + 2^{2}}}

\frac{13}{2 \sqrt{6}}

units.

(ii)
Given planes are,6x + 4y – 12z + 9 = 0 ….(1)

And 3x + 2y – 6z + 1 = 0

So, 6x + 4y – 12z + 2 = 0 …(2)

Since, planes (1) and (2) are parallel so the distance between them is,

d =

\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2} + c^{2}}}

\frac{9 - 2}{\sqrt{6^{2} + 4^{2} + {(- 12)}^{2}}}

\frac{7}{\sqrt{196}}

\frac{1}{2}

units.

13.Please copy the question from book.

Soln:

Let the plane cut the axes at A(a,0,0), B(0,b,0) and C(0,0,C) so that the plane ABC is,

Or,

\frac{x}{a} + \frac{y}{b} + \frac{z}{c}

= 1….(1)

Since, the length of the perpendicular from the origin on the plane (1) is 3p so,

Or, 3p =

\frac{| \frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1 |}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}}

\frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}}

So,

\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}

\frac{1}{9 p^{2}}

…(2)

Let P (α,β,γ) be the centroid of

Δ

ABC, so α =

\frac{a + 0 + 0}{3}

= 3α.

Simi;ar;y, b = 3β and c = 3γ.

Thus, from (2), we get,

\frac{1}{9 a^{2}} + \frac{1}{9 β^{2}} + \frac{1}{9 γ^{2}} = \frac{1}{9 p^{2}}

So,

\frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{γ^{2}} = \frac{1}{p^{2}} .

So, the locus of (α,β,γ) is,

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{1}{p^{2}}

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{1}{p^{2}}

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{1}{p^{2}}

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{1}{p^{2}}

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{1}{p^{2}}

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{1}{p^{2}}

If you find any mistakes then comment in the comment section below.

Mathematics Answer

Unordered List

Friday, 2 December 2016

Chapter 6-Plane Exercise 6.1

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