Question

If the distance between planes $4x - 2y - 4z + 1 = 0$ and $4x - 2y - 4z + d = 0$ is 7, then d is
A.$41\,\,or\,\,- 42$
B. $42\,\, or\,\, - 43$
C. $- 41\,\, or\,\, 43$
D. $- 42\,\, or\,\, 44$

Answer 1

Here we have equations of two planes which are seen to be parallel because they have same coefficients of all variables and only d is different. Using the formula of distance between two planes and substituting the values in the formula we equate it to 7.

Formula used: Formula for distance between two parallel planes $ax + by + cz + {d_1} = 0$ and $ax + by + cz + {d_2} = 0$ is given by $\dfrac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

Complete step-by-step answer:
We have equations of two planes as $4x - 2y - 4z + 1 = 0$ and $4x - 2y - 4z + d = 0$.
Coefficient of x in both equations is 4.
Coefficient of y in both equations is -2.
Coefficient of z in both equations is -4.
Therefore, we can say that two planes are parallel to each other.
Now we compare the equation of plane to general equation of plane $ax + by + cz + {d_1} = 0$
We get $a = 4,b = - 2,c = - 4,{d_1} = 1,{d_2} = d$
We know that distance between two parallel planes is given by $\dfrac{{\left| {{d_1} - {d_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
Substitute the values in the formula.
Distance $= \dfrac{{\left| {1 - d} \right|}}{{\sqrt {{{(4)}^2} + {{( - 2)}^2} + {{( - 4)}^2}} }}$
Squaring the values in the denominator and substituting the value 7 to the distance in LHS

$$\Rightarrow 7 = \dfrac{{\left| {1 - d} \right|}}{{\sqrt {16 + 4 + 16} }} \\\ \Rightarrow 7 = \dfrac{{\left| {1 - d} \right|}}{{\sqrt {36} }} \\\$$

Write the term in the denominator as square of a number
$\Rightarrow 7 = \dfrac{{\left| {1 - d} \right|}}{{\sqrt {{6^2}} }}$
Cancel square root with square power
$\Rightarrow 7 = \dfrac{{\left| {1 - d} \right|}}{6}$
Cross multiply the values

$$\Rightarrow 7 \times 6 = \left| {1 - d} \right| \\\ \Rightarrow 42 = \left| {1 - d} \right| \\\$$

Now we can open modulus in two ways
Case1: when $d < 0$

$$\Rightarrow \left| {1 - d} \right| = - (1 - d) \\\ \Rightarrow \left| {1 - d} \right| = - 1 + d \\\$$

So, $\Rightarrow 42 = - 1 + d$
Shift constant to one side of the equation.

$$\Rightarrow 42 + 1 = d \\\ \Rightarrow 43 = d \\\$$

Case2: when $d \geqslant 0$

$$\Rightarrow \left| {1 - d} \right| = (1 - d) \\\ \Rightarrow \left| {1 - d} \right| = 1 - d \\\$$

So, $\Rightarrow 42 = 1 - d$
Shift constant to one side of the equation.

$$\Rightarrow 42 - 1 = - d \\\ \Rightarrow 41 = - d \\\$$

Multiply both sides by -1

$$\Rightarrow 41 \times - 1 = - d \times - 1 \\\ \Rightarrow - 41 = d \\\$$

Therefore, two values of d are $- 41\,\,or\,\,43$.

So, the correct answer is “Option C”.

Note: Students many times try to find a point on one plane and then find distance from that point to the other plane using the different formula which will be of no help here, instead our solution will become more complex. Keep in mind to always check before solving if the equations of the plane are parallel or not.