Q.13. If the point of intersection of the lines and lies on xy-plane, then the value of is :

Q: What is the correct answer for JEE Main Mathematics Q13?

The correct answer is C. Sol. Let the first line be $L_1$: $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} = \lambda_1$. Any point on $L_1$ is $P(2+\lambda_1, b+4\lambda_1, 2a+7\lambda_1)$. Let the second line be $L_2$: $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} = \lambda_2$. Any point on $L_2$ is $Q(-1+3\lambda_2, -a+5\lambda_2, -b-1+7\lambda_2)$. If the lines intersect, then $P=Q$. $2+\lambda_1 = -1+3\lambda_2 \

Q: Q.13. If the point of intersection of the lines and lies on xy-plane, then the value of is :

The correct answer is C. Sol. Let the first line be $L_1$: $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} = \lambda_1$. Any point on $L_1$ is $P(2+\lambda_1, b+4\lambda_1, 2a+7\lambda_1)$. Let the second line be $L_2$: $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} = \lambda_2$. Any point on $L_2$ is $Q(-1+3\lambda_2, -a+5\lambda_2, -b-1+7\lambda_2)$. If the lines intersect, then $P=Q$. $2+\lambda_1 = -1+3\lambda_2 \

Question

Q.13. If the point of intersection of the lines  and  lies on xy-plane, then the value of  is :

FirstInTest · Accepted Answer

Sol. Let the first line be $L_1$: $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} = \lambda_1$.
Any point on $L_1$ is $P(2+\lambda_1, b+4\lambda_1, 2a+7\lambda_1)$.

Let the second line be $L_2$: $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} = \lambda_2$.
Any point on $L_2$ is $Q(-1+3\lambda_2, -a+5\lambda_2, -b-1+7\lambda_2)$.

If the lines intersect, then $P=Q$.
$2+\lambda_1 = -1+3\lambda_2 \Rightarrow \lambda_1 - 3\lambda_2 = -3$ (1)
$b+4\lambda_1 = -a+5\lambda_2 \Rightarrow 4\lambda_1 - 5\lambda_2 = -a-b$ (2)
$2a+7\lambda_1 = -b-1+7\lambda_2 \Rightarrow 7\lambda_1 - 7\lambda_2 = -b-1-2a$ (3)

The point of intersection lies on the xy-plane, which means its z-coordinate is 0.
So, $2a+7\lambda_1 = 0$ (from $L_1$) and $-b-1+7\lambda_2 = 0$ (from $L_2$).
From $2a+7\lambda_1 = 0 \Rightarrow 7\lambda_1 = -2a$.
From $-b-1+7\lambda_2 = 0 \Rightarrow 7\lambda_2 = b+1$.

Substitute $7\lambda_1 = -2a$ into (3): $-2a - 7\lambda_2 = -b-1-2a \Rightarrow -7\lambda_2 = -b-1 \Rightarrow 7\lambda_2 = b+1$. This is consistent.

Now we have $\lambda_1 = -\frac{2a}{7}$ and $\lambda_2 = \frac{b+1}{7}$.
Substitute these into equation (1):
$-\frac{2a}{7} - 3\left(\frac{b+1}{7}\right) = -3$
$-2a - 3(b+1) = -21$
$-2a - 3b - 3 = -21$
$-2a - 3b = -18$
$2a + 3b = 18$ (4)

Now substitute $\lambda_1$ and $\lambda_2$ into equation (2):
$4\left(-\frac{2a}{7}\right) - 5\left(\frac{b+1}{7}\right) = -a-b$
$-\frac{8a}{7} - \frac{5b+5}{7} = -a-b$
$-8a - 5b - 5 = -7a - 7b$
$-8a + 7a - 5b + 7b = 5$
$-a + 2b = 5$ (5)

We have a system of two linear equations for $a$ and $b$:
1) $2a + 3b = 18$
2) $-a + 2b = 5$

From (5), $a = 2b - 5$.
Substitute this into (4):
$2(2b - 5) + 3b = 18$
$4b - 10 + 3b = 18$
$7b = 28$
$b = 4$.

Now find $a$:
$a = 2(4) - 5 = 8 - 5 = 3$.

So, $a=3$ and $b=4$.
The value of $a+b = 3+4 = 7$.

Answer

2

Answer

5

Answer

7

Answer

9

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