Let the points ( 1 , 1 , p ) and ( - 3 , 0 , 1 ) be equidistant from the plane r ? · ( 3 i + 4 j - 12 k ) + 13 = 0 , then the value of p is:

Question

Let the points $(1, 1, p)$ and $(- 3, 0, 1)$ be equidistant from the plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0,$ then the value of $p$ is:

Smart Achievers · Accepted Answer

(2)

Ans. $\frac{7}{3}$

Explanation:

The distance of point $(1, 1, p)$ from the plane

$\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0$

or (in Cartesian form) $3 x + 4 y - 12 z + 13 = 0$ is

$d_{1} =$ $| \frac{3 \times 1 + 4 \times 1 - 12 \times p + 13}{\sqrt{3^{2} + 4^{2} + {(- 12)}^{2}}} |$

$=$ $| \frac{3 + 4 - 12 p + 13}{\sqrt{169}} |$ $=$ $| \frac{20 - 12 p}{13} |$ $\dots (i)$

The distance of the point $(- 3, 0, 1)$ from the plane $3 x + 4 y - 12 z + 13 = 0$ is

$d_{2} =$ $| \frac{3 \times (- 3) + 4 \times 0 - 12 \times 1 + 13}{\sqrt{3^{2} + 4^{2} + {(- 12)}^{2}}} |$

$= \frac{8}{13}$

According to the given condition,

$d_{1} = d_{2}$

$\Rightarrow$ $| \frac{20 - 12 p}{13} |$ $= \frac{8}{13}$

$\Rightarrow$ $\frac{20 - 12 p}{13} = \pm \frac{8}{13}$

Taking +ve sign, we get

$\frac{20 - 12 p}{13} = \frac{8}{13}$

$\Rightarrow$ $p = 1$

Taking $-$ ve sign, we get

$\frac{20 - 12 p}{13} = - \frac{8}{13}$

$\Rightarrow$ $p = \frac{28}{12} = \frac{7}{3}$

Solution

Q.
Let the points $(1, 1, p)$ and $(- 3, 0, 1)$ be equidistant from the plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0,$ then the value of $p$ is:

1 $\frac{3}{7}$

2 $\frac{7}{3}$

3 $\frac{4}{3}$

4 $\frac{3}{4}$

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Solution

Q. Let the points (1,1,p) and (-3,0,1) be equidistant from the plane r→·(3i^+4j^-12k^)+13=0, then the value of p is:

1 37

2 73

3 43