Download App

Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsVectors2 Marks
MCQ
If line joining points $A$ and $B$ having position vectors $6 \vec{a}-4 \vec{b}+4 \vec{c}$ and $-4 \vec{c}$ respectively, and the line joining the points $C$ and $D$ having position vectors $-\vec{a}-2 \vec{b}-3 \vec{c}$ and $\vec{a}+2 \vec{b}-5 \vec{c}$ intersect, then their point of intersection is
  • ✓
    $B$
  • B
    $C$
  • C
    $D$
  • D
    $A$
✓

Answer

Correct option: A.
$B$
(a) : P.V.of $A, B, C$, and $D$ are $6 \vec{a}-4 \vec{b}+4 \vec{c}$, $-4 \vec{c},-\vec{a}-2 \vec{b}-3 \vec{c}$ and $\vec{a}+2 \vec{b}-5 \vec{c}$ respectively
$
\begin{aligned}
& \Rightarrow \overrightarrow{A B}=-6 \vec{a}+4 \vec{b}-8 \vec{c} \text { and } \overrightarrow{C D}=2 \vec{a}+4 \vec{b}-2 \vec{c} \\
& \therefore \quad \vec{r}=(6 \vec{a}-4 \vec{b}+4 \vec{c})+\lambda(-6 \vec{a}+4 \vec{b}-8 \vec{c}) \\
& \text { and } \vec{r}=(-\vec{a}-2 \vec{b}-3 \vec{c})+\mu(2 \vec{a}+4 \vec{b}-2 \vec{c}) \\
& \Rightarrow(6-6 \lambda)=(-1+2 \mu) \\
&(-4+4 \lambda)=(-2+4 \mu) \\
&(4-8 \lambda)=(-3-2 \mu)
\end{aligned}
$
On solving (i), (ii) and (iii), we get
$
\lambda=1, \mu=1 / 2
$
So the point of intersection is :
$
\begin{aligned}
& (6 \vec{a}-4 \vec{b}+4 \vec{c})+1(-6 \vec{a}+4 \vec{b}-8 \vec{c}) \\
& =6 \vec{a}-4 \vec{b}+4 \vec{c}-6 \vec{a}+4 \vec{b}-8 \vec{c}=-4 \vec{c} \text { i.e., } B
\end{aligned}
$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Vectors→Vectors — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $X$ is a random variable with probability mass function
$P (x)=k x \quad, \quad$ for $x=1,2,3$
$=0$. otherwise then $k=\ldots$
→
Let [x] denote the greatest integer less than or equal to $x$. If $f(x) = \sin^{-1}x$, $g(x) = [x^2]$ and $\text{h(x)}=2\text{x},\frac{1}{2}\leq\text{x}\leq\frac{1}{\sqrt{2}},$ then
→
If $|\bar{a}|=3,|\bar{b}|=4$ and the angle between $\bar{a}$ and $\overline{ b }$ is $120^{\circ}$, then $|4 \overline{ a }+3 \overline{b}|$ is equal to
→
The angle between the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$ and $\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$ is:
→
If $\int_0^1 \frac{ d x}{\sqrt{1+x}-\sqrt{x}}=\frac{ k }{3}$, then $k$ is equal to_____.
→
D.c.s of a line segment AB are $\frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$, $\frac{-2}{\sqrt{17}}$. If $AB =\sqrt{17}$ and $A \equiv(3,-6,10)$, then co-ordinates of B will be
→
$\int \frac{1}{x} \cdot \log x d x= $______.
→
The equation of the normal to the curve $y = x(2 - x)$ at the point $(2, 0)$ is:
→
The acute angle between the lines represented by $6 x^2-x y-y^2=0$ is
→
$\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{54}}\text{ dx}=$
→