JEE Advanced · Mathematics · 31. 3D Geometry

Three lines
$L_{1} : \vec{r} = λ \hat{i}, λ \in R,$
$L_{2} : \vec{r} = \hat{k} + μ \hat{j}, μ \in R$ and
$L_{3} : \vec{r} = \hat{i} + \hat{j} + ν \hat{k}, ν \in R$
are given. For which point(s) $Q$ and $L_{2}$ can we find a point $P$ on $L_{1}$ and a point $R$ on $L_{3}$ so that $P, Q$ and $R$ are collinear?

A $\hat{k} + \hat{j}$
B $\hat{k}$
C $\hat{k} + \frac{1}{2} \hat{j}$
D $\hat{k} - \frac{1}{2} \hat{j}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\hat{k} - \frac{1}{2} \hat{j}$

Step-by-step Solution

Detailed explanation

As given

L_{1} \Rightarrow \vec{r} = λ i, λ \in R

L_{2} \Rightarrow \vec{r} = k + μ j, μ \in R

and

L_{3} \Rightarrow \vec{r} = i + j + ν k, ν \in R

Let point

P (λ, 0, 0), Q (0, μ, 1)

and

R (1, 1, ν)

and

\vec{P Q} = - λ \hat{i} + μ \hat{j} + \hat{k}, \vec{P R} = (1 - λ) \hat{i} + \hat{j} + ν \hat{k}

P, Q

and

R

are co-linear

\Rightarrow \vec{P Q}

is parallel to

\vec{P R}

\Rightarrow - \frac{λ}{1 - λ} = \frac{μ}{1} = \frac{1}{ν}

\Rightarrow λ = \frac{μ}{μ - 1}

and

ν = \frac{1}{μ}

hence

\begin{matrix} μ \neq 0 & μ \neq 1 \end{matrix}

Q \neq (0, 0, 1)

and

Q \neq (0, 1, 1)

Q \neq \hat{k} & Q \neq (\hat{j} + \hat{k})

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Column I		Column II
$(A)$	In a triangle $∆ X Y Z$ let $a, b$ and $c$ be the lengths of the angles $X, Y$ and $Z$ , respectively. If $2 (a^{2} - b^{2}) = c^{2}$ and $λ = \frac{\sin (X - Y)}{sinZ}$ then possible values of $n$ for which $\cos (n π λ) = 0$ is are	$(P)$	$1$
$(B)$	In a triangle $Δ X Y Z,$ let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$ , respectively. If $1 + \cos 2 x - 2 \cos 2 y$ $= 2 \sin x \sin y$ , then possible value(s) of $\frac{a}{b}$ is (are)	$(Q)$	$2$
$(C)$	In $R^{2},$ let $\sqrt{3} \hat{i} + \hat{j}, \hat{i} + \sqrt{3} \hat{j}$ and $β \hat{i} + (1 - β) \hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$ , respectively. If the distance of $Z$ from the bisector of the acute angle of $\vec{O X}$ with $\vec{O Y}$ is $\frac{3}{\sqrt{2}}$ , then possible value (s) of $\|β\|$ is (are)	$(R)$	$3$
$(D)$	Suppose that $F (α)$ denotes the area of the region bounded by $x = 0, x = 2, y^{2} = 4 x$ and $y = \|a x - 1\| +$ $\|α x - 2\| + α x,$ where $α \in \{0, 1\} .$ Then the value (s) of $F (α) + \frac{8}{3} \sqrt{2},$ when $α = 0$ and $α = 1$ , is (are)	$(S)$	$5$
		$(T)$	$6$

JEE Advanced 2015 Hard

Paragraph:
Read the following passage and answer the questions.
For every function $f(x)$ which is twice differentiable, these will be good approximation of $\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}$. Now, if we take $c=\frac{a+b}{2}$, then using above again, we get $\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}$ and so on.
We get approximation for value of $\int_a^b f(x) d x$.Question:
If $f^{\prime \prime}(x) < 0, \forall x \in(a, b), c(c, f(c))$ is point of maxima where $c \in(a, b)$, then $f^{\prime}(c)$ is

JEE Advanced 2006 Easy

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Three linesL1:r→=λi^, λ∈R,L2:r→=k^+μj^, μ∈R andL3: r→=i^+j^+νk^, ν∈Rare given. For which point(s) Q and L2 can we find a point P on L1 and a point R on L3 so that P, Q and R are collinear?

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