1/3, 1/6, 1

Hint:

Given 3x + 1 = 6y – 2 = 1 – z

= (3x + 1)/1 = (6y – 2)/1 = (1 – z)/1

= (x + 1/3)/(1/3) = (y – 2/6)/(1/6) = (1 – z)/1

= (x + 1/3)/(1/3) = (y – 1/3)/(1/6) = (1 – z)/1

Now, the direction ratios are: 1/3, 1/6, 1

(-3, 5, 2)

Hint:

Let image of the point P(1, 3, 4) is Q in the given plane.

The equation of the line through P and normal to the given plane is

(x-1)/2 = (y-3)/-1 = (z-4)/1

Since the line passes through Q, so let the coordinate of Q are (2r + 1, -r + 3, r + 4)

Now, the coordinate of the mid-point of PQ is

(r + 1, -r/2 + 3, r/2 + 4)

Now, this point lies in the given plane.

2(r + 1) – (-r/2 + 3) + (r/2 + 4) + 3 = 0

⇒ 2r + 2 + r/2 – 3 + r/2 + 4 + 3 = 0

⇒ 3r + 6 = 0

⇒ r = -2

Hence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)

= (-3, 5, 2)

meet in a unique point

Hint:

Given, three planes are

x + y = 0 …….. 1

y + z = 0 …….. 2

and x + z = 0 ……… 3

add these planes, we get

2(x + y + z) = 0

⇒ x + y + z = 0 ……… 4

From equation 1

0 + z = 0

⇒ z = 0

From equation 2

x + 0 = 0

⇒ x = 0

From equation 3

y + 0 = 0

⇒ y = 0

So, (x, y, z) = (0, 0, 0)

Hence, the three planes meet in a unique point.

(5/3, 7/3, 17/3)

Hint:

Let D be the foot of perpendicular and let it divide BC in the ration m : 1

Then the coordinates of D are {(3m + 4)/(m + 1), (5m + 7)/(m + 1), (3m + 1)/(m + 1)}

Now, AD ⊥ BC

⇒ AD . BC = 0

⇒ -(2m + 3) – 2(5m + 7) – 4 = 0

⇒ m = -7/4

So, the coordinate of D are (5/3, 7/3, 17/3)

Plane

Hint:

Let the position vectors of the given points A and B be a and b respectively and that of the variable point be r.

Now, given that

PA² – PB² = k (constant)

⇒ |AP|² – |BP|² = k

⇒ |r – a|² – |r – b|² = k

⇒ (|r|² + |a|² – 2r.a) – (|r|² + |b|² – 2r.b) = k

⇒ 2r.(b – a) = k + |b|² – |a|²

⇒ r.(b – a) = (k + |b|² – |a|²)/2

⇒ r.(b – a) = C where C = (k + |b|² – |a|²)/2 = constant

So, it represents the equation of a plane.

9x² + 25y² + 25z² – 225 = 0

Hint:

Let the point P is (x, y, z)

Now given that

PA + PB = 10

⇒ √{(x-4)² + y² + z²} + √{(x+4)² + y² + z²} = 10

⇒ √{(x-4)² + y² + z²} = 10 – √{(x+4)² + y² + z²}

Now square both side

[√{(x-4)² + y² + z²}]² = (10)² + [{(x+4)² + y² + z²}]² – 2 ×10×√{(x+4)² + y² + z²}

⇒ {(x-4)² + y² + z²} = 100 + {(x+4)² + y² + z²} – 20×√{(x+4)² + y² + z²}

⇒ x² + 16 – 8x + y² + z² = 100 + x² + 16 + 8x + y² + z² – 20×√{(x+4)² + y² + z²}

⇒ – 8x = 100 + 8x – 20×√{(x+4)² + y² + z²}

⇒ -8x -8x – 100 = – 20×√{(x+4)² + y² + z²}

⇒ -16x -100 = – 20×√{(x+4)² + y² + z²}

⇒ 4x + 25 = 5×√{(x+4)² + y² + z²}

Again square both side,

(4x + 25)² = 25 ×[√{(x+4)² + y² + z²}]²

⇒ 16x² + 625 + 200x = 25×{(x+4)² + y² + z²}

⇒ 16x² + 625 + 200x = 25×(x² + 16 + 8x + y² + z²)

⇒ 16x² + 625 + 200x = 25x² + 400 + 200x + 25y² + 25z²

⇒ 25x² + 400 + 200x + 25y² + 25z² – 16x² – 625 – 200x = 0

⇒ 9x² + 25y² + 25z² – 225 = 0

5

Hint:

Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)

Now distance = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}

⇒ distance = √{(4cos θ – 3sin θ)²}

⇒ distance = 4cos θ – 3sin θ ……………. 1

Now, maximum value of 4cos θ – 3sin θ = √{(4² + (-3)²}

= √(16 + 9)

= √25

= 5

From equation 1, we get

distance = 5

So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5

2√3(i + j + k)

Hint:

Let l, m, n are DCs of r.

Given, l = m = n

⇒ l² + m² + n² = 1

⇒ 3l² = 1

⇒ l² = 1/3

⇒ l = m = n = 1/√3

So, r = |r|(li + mj + nk)

⇒ r = 6(i/√3 + j/√3 + k/√3)

⇒ r = 2√3(i + j + k)

a

Hint:

Given, equation of plane is:

2x – (1 + a)y + 3az = 0

=> (2x – y) + a(-y + 3z) = 0

which is passing through the intersection of the planes

2x – y = 0 and -y + 3z = 0

2x – y = 0 and y – 3z = 0

√3 unit

Hint:

Let a is the length of the side of a square.

Given, the diagonal of a square are (1,–2,3) and (2, -3, 5)

Now, length of the diagonal of square = √{(1 – 2)² + (-2 + 3)² + (3 – 5)²}

= √{1 + 1 + 4}

= √6

Again length of the diagonal of square is √2 times the length of side of the square.

⇒ a√2 = √6

⇒ a√2 = √3×√2

⇒ a = √3

So, the length of side of square is √3 unit

(0, 17/2, -13/2)

Hint:

The line passing through the points (5, 1, 6) and (3, 4, 1) is given as

(x-5)/(3-5) = (y-1)/(4-1) = (z-6)/(1-6)

⇒ (x-5)/(-2) = (y-1)/3 = (z-6)/(-5) = k(say)

⇒ (x-5)/(-2) = k

⇒ x – 5 = -2k

⇒ x = 5 – 2k

(y-1)/3 = k

⇒ y – 1 = 3k

⇒ y = 3k + 1

and (z-6)/(-5) = k

⇒ z – 6 = -5k

⇒ z = 6 – 5k

Now, any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k)

The equation of YZ-plane is x = 0

Since the line passes through YZ-plane

So, 5 – 2k = 0

⇒ k = 5/2

Now, 3k + 1 = 3 × 5/2 + 1 = 15/2 + 1 = 17/2

and 6 – 5k = 6 – 5×5/2 = 6 – 25/2 = -13/2

Hence, the required point is (0, 17/2, -13/2)

 π/3

Hint:

Let a is a vector parallel to the vector having direction ratio is 4, -3, 5

⇒ a = 4i – 3j + 5k

Let b is a vector parallel to the vector having direction ratio is 3 ,4, 5

⇒ b = 3i + 4j + 5k

Let θ be the angle between the given vectors.

Now, cos θ = (a . b)/(|a|×|b|)

⇒ cos θ = (12 – 12 + 25)/{√(16 + 9 + 25)×√(9 + 16 + 25)}

⇒ cos θ = 25/{√(50)×√(50)}

⇒ cos θ = 25/50

⇒ cos θ = 1/2

⇒ cos θ = π/3

⇒ θ = π/3

So, the angle between the vectors with direction ratios are 4, -3, 5 and 3, 4, 5 is π/3

 r . (2i – j + 2k) = 3

Hint:

The equation of plane parallel to the plane r . (2i – j + 2k) = 5 is

r . (2i – j + 2k) = d

Since it passes through the point i + j + k, therefore

(i + j + k) . (2i – j + 2k) = d

⇒ d = 2 – 1 + 2

⇒ d = 3

So, the required equation of the plane is

r . (2i – j + 2k) = 3

either (0, -1, 0) or (0, 7, 0)

Hint:

Let the point on y-axis is O(0, y, 0)

Given point is A(2, 3, -1)

Given OA = 3

⇒ OA² = 9

⇒ (2 – 0)² + (3 – y)² + (-1 – 0)² = 9

⇒ 4 + (3 – y)² + 1 = 9

⇒ 5 + (3 – y)² = 9

⇒ (3 – y)² = 9 – 5

⇒ (3 – y)² = 4

⇒ 3 – y = √4

⇒ 3 – y = ±4

⇒ 3 – y = 4 and 3 – y = -4

⇒ y = -1, 7

So, the point is either (0, -1, 0) or (0, 7, 0)

None of these

Hint:

Let l, m, n be the direction cosines of the given vector.

Then, α, β, γ

l = cos α

m = cos β

n = cos γ

Now, l² + m² + n² = 1

⇒ cos² α + cos² β + cos² γ = 1

⇒ 1 – sin² α + 1 – sin² β + 1 – sin² γ = 1

⇒ 3 – sin² α – sin² β – sin² γ = 1

⇒ 3 – 1 = sin² α + sin² β + sin² γ

⇒ sin² α + sin² β + sin² γ = 2

3 : 2

Hint:

Since OP has projections 13/5, 19/5 and 26/5 on the coordinate axes, therefore

OP = 13i/5 + 19j/5 + 26/5k

Let P divides the join of Q(2, 2, 4) and R(3, 5, 6) in the ratio m : 1

Then the position vector of P is

{(3m + 2)/(m + 1), (5m + 2)/(m + 1), (6m + 4)/(m + 1)}

So, 13i/5 + 19j/5 + 26/5k = (3m + 2)/(m + 1)+ (5m + 2)/(m + 1)+ (6m + 4)/(m + 1)

⇒ (3m + 2)/(m + 1) = 13/5

⇒ 2m = 3

⇒ m = 3/2

⇒ m : 1 = 3 : 2

Hence, P divides QR in the ration 3 : 2

a plane containing Z axis

Hint:

Given, equation is 3x – 4y = 0

Here z = 0

So, the given equation 3x – 4y = 0 represents a plane containing Z axis.