Question

The distance of the point $(4,2,3)$ from the plane $\vec{r}\cdot(6\hat{i}+2\hat{j}-9\hat{k})=46$ is

• A. $\frac{23}{5}$
• B. $\frac{46}{11}$
• C. $\frac{45}{11}$
• D. $\frac{11}{45}$
• E. $\frac{5}{23}$

Hint:

Formula Tip: The denominator is simply the magnitude of the normal vector. For vectors like $(6, 2, -9)$, recognizing standard Pythagorean quadruples (like $36+4+81=121=11^2$) speeds up calculation.

Answer 1

The Correct Option is C

Concept:
The perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to a plane given by the Cartesian equation $ax + by + cz + d = 0$ is calculated using the distance formula: $$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Step 1: Convert the plane to Cartesian form.
The given vector equation is $\vec{r} \cdot (6\hat{i} + 2\hat{j} - 9\hat{k}) = 46$. Substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ to get the Cartesian equation: $$6x + 2y - 9z = 46$$ Bring the constant to the left side: $$6x + 2y - 9z - 46 = 0$$

Step 2: Identify the parameters for the formula.
From the plane equation: $a = 6$, $b = 2$, $c = -9$, and $d = -46$. The given point is $(x_1, y_1, z_1) = (4, 2, 3)$.

Step 3: Calculate the numerator (absolute value).
Substitute the point into the plane's expression: $$|ax_1 + by_1 + cz_1 + d| = |6(4) + 2(2) - 9(3) - 46|$$ $$= |24 + 4 - 27 - 46|$$ $$= |28 - 73| = |-45| = 45$$

Step 4: Calculate the denominator (magnitude of normal).
Find the length of the normal vector $\sqrt{a^2 + b^2 + c^2}$: $$= \sqrt{6^2 + 2^2 + (-9)^2}$$ $$= \sqrt{36 + 4 + 81}$$ $$= \sqrt{121} = 11$$

Step 5: Determine the final distance.
Divide the numerator by the denominator: $$d = \frac{45}{11}$$ Hence the correct answer is (C) $\frac{45{11}$}.