Q.15. The square of the distance of the point of intersection of the lines , and from the origin i

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

The correct answer is C. Sol. Equating both the equations: $(\mathbf{i}+\mathbf{j}-\mathbf{k})+\lambda(2\mathbf{a}\mathbf{i}-\mathbf{j}) = (4\mathbf{i}-\mathbf{k})+\mu(2\mathbf{i}+\mathbf{a}\mathbf{k})$ Comparing components: For $\mathbf{i}$: $1 + 2\lambda\mathbf{a} = 4 + 2\mu$ (Equation 1) For $\mathbf{j}$: $1 - \lambda = 0 \Rightarrow \lambda = 1$ (Equation 2) For $\mathbf{k}$: $-1 = -1 + \mu\mathbf{a} \Rightarrow \mu\m

Q: Q.15. The square of the distance of the point of intersection of the lines , and from the origin is :

The correct answer is C. Sol. Equating both the equations: $(\mathbf{i}+\mathbf{j}-\mathbf{k})+\lambda(2\mathbf{a}\mathbf{i}-\mathbf{j}) = (4\mathbf{i}-\mathbf{k})+\mu(2\mathbf{i}+\mathbf{a}\mathbf{k})$ Comparing components: For $\mathbf{i}$: $1 + 2\lambda\mathbf{a} = 4 + 2\mu$ (Equation 1) For $\mathbf{j}$: $1 - \lambda = 0 \Rightarrow \lambda = 1$ (Equation 2) For $\mathbf{k}$: $-1 = -1 + \mu\mathbf{a} \Rightarrow \mu\m

Q: What are the options for this JEE Main Mathematics question?

A. 5 · B. 10 · C. 17 · D. 26

Question

Q.15. The square of the distance of the point of intersection of the lines ,  and  from the origin is :

FirstInTest · Accepted Answer

Sol. Equating both the equations:
$(\mathbf{i}+\mathbf{j}-\mathbf{k})+\lambda(2\mathbf{a}\mathbf{i}-\mathbf{j}) = (4\mathbf{i}-\mathbf{k})+\mu(2\mathbf{i}+\mathbf{a}\mathbf{k})$
Comparing components:
For $\mathbf{i}$: $1 + 2\lambda\mathbf{a} = 4 + 2\mu$ (Equation 1)
For $\mathbf{j}$: $1 - \lambda = 0 \Rightarrow \lambda = 1$ (Equation 2)
For $\mathbf{k}$: $-1 = -1 + \mu\mathbf{a} \Rightarrow \mu\mathbf{a} = 0$. Since $\mathbf{a} 
e 0$, we must have $\mu = 0$ (Equation 3).
Substitute $\lambda=1$ and $\mu=0$ into Equation 1:
$1 + 2(1)\mathbf{a} = 4 + 2(0)$
$1 + 2\mathbf{a} = 4 \Rightarrow 2\mathbf{a} = 3 \Rightarrow \mathbf{a} = \frac{3}{2}$.
Now find the point of intersection using $\lambda=1$ in the first line equation:
$\vec{r} = (\mathbf{i}+\mathbf{j}-\mathbf{k})+1(2\mathbf{a}\mathbf{i}-\mathbf{j}) = \mathbf{i}+\mathbf{j}-\mathbf{k}+2\mathbf{a}\mathbf{i}-\mathbf{j} = (1+2\mathbf{a})\mathbf{i} - \mathbf{k}$.
Substitute $\mathbf{a} = \frac{3}{2}$:
$\vec{r} = (1+2(\frac{3}{2}))\mathbf{i} - \mathbf{k} = (1+3)\mathbf{i} - \mathbf{k} = 4\mathbf{i} - \mathbf{k}$.
The point of intersection is $(4, 0, -1)$.
The square of the distance from the origin $(0,0,0)$ is $4^2 + 0^2 + (-1)^2 = 16 + 0 + 1 = 17$.

Answer

5

Answer

10

Answer

17

Answer

26

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