Q.15. Let the image of the point in the line , , be . If , , is the point on such that the distanc

Q: What is the correct answer for JEE Main Q15?

The correct answer is C. Sol. Let $Q$ be the foot of the perpendicular from $P(1,6,a)$ to the line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$. Let $Q = (k, 2k+1, bk+a-1)$. Direction vector of $L$ is $\vec{v} = (1,2,b)$. Vector $\vec{PQ} = (k-1, 2k+1-6, bk+a-1-a) = (k-1, 2k-5, bk-1)$. Since $\vec{PQ} \perp \vec{v}$, $\vec{PQ} \cdot \vec{v} = 0$. $(k-1)(1) + (2k-5)(2) + (bk-1)(b) = 0$ $k-1 + 4k-10 + b^2k-b = 0$ $(5+

Q: Q.15. Let the image of the point in the line , , be . If , , is the point on such that the distance of from the foot of perpendicular from the point on is , then is equal to:

The correct answer is C. Sol. Let $Q$ be the foot of the perpendicular from $P(1,6,a)$ to the line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$. Let $Q = (k, 2k+1, bk+a-1)$. Direction vector of $L$ is $\vec{v} = (1,2,b)$. Vector $\vec{PQ} = (k-1, 2k+1-6, bk+a-1-a) = (k-1, 2k-5, bk-1)$. Since $\vec{PQ} \perp \vec{v}$, $\vec{PQ} \cdot \vec{v} = 0$. $(k-1)(1) + (2k-5)(2) + (bk-1)(b) = 0$ $k-1 + 4k-10 + b^2k-b = 0$ $(5+

Question

Q.15. Let the image of the point  in the line , , be . If , , is the point on  such that the distance of  from the foot of perpendicular from the point  on  is , then  is equal to:

FirstInTest · Accepted Answer

Sol. Let $Q$ be the foot of the perpendicular from $P(1,6,a)$ to the line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$.
Let $Q = (k, 2k+1, bk+a-1)$.
Direction vector of $L$ is $\vec{v} = (1,2,b)$.
Vector $\vec{PQ} = (k-1, 2k+1-6, bk+a-1-a) = (k-1, 2k-5, bk-1)$.
Since $\vec{PQ} \perp \vec{v}$, $\vec{PQ} \cdot \vec{v} = 0$.
$(k-1)(1) + (2k-5)(2) + (bk-1)(b) = 0$
$k-1 + 4k-10 + b^2k-b = 0$
$(5+b^2)k - (11+b) = 0$ ...(A)
$Q$ is the midpoint of $PP'$. $P'(3/2, 0, a+c)$.
$Q = (\frac{1+3/2}{2}, \frac{6+0}{2}, \frac{a+a+c}{2}) = (\frac{5}{4}, 3, a+\frac{c}{2})$.
Comparing coordinates of $Q$: $k = \frac{5}{4}$, $2k+1 = 3 \Rightarrow 2(\frac{5}{4})+1 = \frac{5}{2}+1 = \frac{7}{2} 
e 3$. This indicates an error in my interpretation or the problem statement's given image point $P'$.
Let's re-read: "The image of point P(1, 6, a) in the line L is P'(3/2,0,a+c)". This is the image, not the foot of the perpendicular.
Let $Q(x_Q, y_Q, z_Q)$ be the foot of the perpendicular from $P$ to $L$.
$Q$ is the midpoint of $PP'$.
$x_Q = \frac{1+3/2}{2} = \frac{5/2}{2} = \frac{5}{4}$
$y_Q = \frac{6+0}{2} = 3$
$z_Q = \frac{a+a+c}{2} = a+\frac{c}{2}$
So $Q = (\frac{5}{4}, 3, a+\frac{c}{2})$.
Since $Q$ lies on $L$, we have $\frac{5/4}{1} = \frac{3-1}{2} = \frac{a+c/2-a+1}{b}$
$\frac{5}{4} = \frac{2}{2} = 1$. This is a contradiction: $\frac{5}{4} 
e 1$. This means the given $P'$ is not the image in the line $L$ as described by the equation. There must be a typo in the question or the provided solution. Let's follow the provided solution's steps.

**Following the provided solution:**
Sol. Let $Q$ be the point from $P(1,6,a)$ to the line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$.
The image of point $P$ in line $L$ is $P'(\frac{3}{2},0,a+c)$. (The solution uses $P'$ as the image, but then uses $Q$ as the midpoint of $PP'$ which is correct for image definition).
$Q$ is the mid point $PP'$:
$Q = (\frac{1+3/2}{2}, \frac{6+0}{2}, \frac{a+a+c}{2}) = (\frac{5}{4}, 3, a+\frac{c}{2})$.
Comparing $x$ and $y$ coordinates of $Q$ with the line equation $\frac{x}{1} = \frac{y-1}{2} = k$:
$x_Q = k = \frac{5}{4}$.
$y_Q = 2k+1 = 2(\frac{5}{4})+1 = \frac{5}{2}+1 = \frac{7}{2}$.
But $y_Q = 3$. So $\frac{7}{2} = 3$ is a contradiction. The solution provided in the image has a mistake in the calculation of $Q$. Let's assume the question meant $P'(x_0, y_0, z_0)$ is the image and $Q$ is the foot of the perpendicular.

Let's re-evaluate based on the provided solution's calculation for $Q$ and then its subsequent steps, assuming the initial $P'$ was a typo and the solution's $Q$ is correct for the problem.
From the solution, $Q = (1,3, \frac{6+c}{2})$. This $Q$ is derived from $P(1,6,a)$ and $P'(\frac{3}{2},0,a+c)$ as midpoint. This is inconsistent with the $P'$ given. Let's assume the solution's $Q$ is the foot of the perpendicular.

If $Q = (1,3, \frac{6+c}{2})$ is on the line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$, then:
$1/1 = (3-1)/2 = (\frac{6+c}{2}-a+1)/b$
$1 = 1 = (\frac{6+c-2a+2}{2})/b \Rightarrow b = \frac{8+c-2a}{2}$ ...(Eq. 1)

Direction vector of $L$ is $\vec{v} = (1,2,b)$.
Vector $\vec{PQ} = (1-1, 3-6, \frac{6+c}{2}-a) = (0, -3, \frac{6+c-2a}{2})$.
Since $\vec{PQ} \perp \vec{v}$, $\vec{PQ} \cdot \vec{v} = 0$.
$(0)(1) + (-3)(2) + (\frac{6+c-2a}{2})(b) = 0$
$-6 + b(\frac{6+c-2a}{2}) = 0$
Substitute $b = \frac{8+c-2a}{2}$ from Eq. 1:
$-6 + (\frac{8+c-2a}{2})(\frac{6+c-2a}{2}) = 0$
$-6 + \frac{(8+c-2a)(6+c-2a)}{4} = 0$
$24 = (8+c-2a)(6+c-2a)$.
This is getting complicated. Let's follow the solution's steps exactly, assuming there's an implicit correction in the problem statement that makes the solution valid.

Solution's $Q = (1,3, \frac{6+c}{2})$.
From $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b} = k$
$x=k, y=2k+1, z=bk+a-1$
For $Q(1,3, \frac{6+c}{2})$:
$k=1$
$3 = 2(1)+1$, which is true.
$\frac{6+c}{2} = b(1)+a-1 \Rightarrow \frac{6+c}{2} = b+a-1 \Rightarrow 6+c = 2b+2a-2 \Rightarrow c = 2b+2a-8$ ...(Eq. A)

Vector $\vec{PQ} = (1-1, 3-6, \frac{6+c}{2}-a) = (0, -3, \frac{6+c-2a}{2})$.
$\vec{v} = (1,2,b)$.
$\vec{PQ} \cdot \vec{v} = 0 \Rightarrow (0)(1) + (-3)(2) + (\frac{6+c-2a}{2})(b) = 0$
$-6 + b(\frac{6+c-2a}{2}) = 0 \Rightarrow b(6+c-2a) = 12$ ...(Eq. B)
From (Eq. A), $c-2a = 2b-8$. Substitute into (Eq. B):
$b(6 + 2b-8) = 12 \Rightarrow b(2b-2) = 12 \Rightarrow 2b^2-2b-12 = 0 \Rightarrow b^2-b-6 = 0$
$(b-3)(b+2) = 0$. Since $b>0$, $b=3$.
Substitute $b=3$ into (Eq. A): $c = 2(3)+2a-8 = 6+2a-8 = 2a-2$.

Now, the solution states $Q(1,3,5)$. This means $\frac{6+c}{2} = 5 \Rightarrow 6+c=10 \Rightarrow c=4$.
If $c=4$ and $b=3$, then from $c=2a-2$, we get $4=2a-2 \Rightarrow 2a=6 \Rightarrow a=3$.
So $P(1,6,3)$, $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-3+1}{3} = \frac{z-2}{3}$.
$Q(1,3,5)$ is on $L$: $1/1 = (3-1)/2 = (5-2)/3 \Rightarrow 1=1=1$. This is consistent.

Now, any point $S$ on the line $L$ can be written as $(k, 2k+1, 3k+2)$.
Distance of $S$ from $Q$ is $2\sqrt{14}$.
$SQ^2 = (k-1)^2 + (2k+1-3)^2 + (3k+2-5)^2 = (2\sqrt{14})^2$
$(k-1)^2 + (2k-2)^2 + (3k-3)^2 = 4 \cdot 14 = 56$
$(k-1)^2 + 4(k-1)^2 + 9(k-1)^2 = 56$
$14(k-1)^2 = 56$
$(k-1)^2 = 4 \Rightarrow k-1 = \pm 2$
$k-1 = 2 \Rightarrow k=3$
$k-1 = -2 \Rightarrow k=-1$
For $k=3$, $S = (3, 2(3)+1, 3(3)+2) = (3,7,11)$.
For $k=-1$, $S = (-1, 2(-1)+1, 3(-1)+2) = (-1,-1,-1)$.
Given $\alpha > 0$, so we choose $S=(3,7,11)$. Thus $\alpha=3, \beta=7, \gamma=11$.
$\alpha+\beta+\gamma = 3+7+11 = 21$.

This matches the solution's final answer. The initial $P'$ and $Q$ values in the solution were confusing, but the derivation of $a,b,c$ and then $S$ is consistent.

-   `question_text`:

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