Download App

VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA1 Mark
MCQ
Choose the correct answer from the given four options.
The vectors $\lambda\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},\ \hat{\text{i}}+\lambda\hat{\text{j}}-\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$ are coplanar if:
  • ✓
    $\lambda=-2$
  • B
    $\lambda=0$
  • C
    $\lambda=1$
  • D
    $\lambda=-1$
✓

Answer

Correct option: A.
$\lambda=-2$
Let $\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}=\hat{\text{i}}+\lambda\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$

For $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ to be coplanar,

$\begin{vmatrix}\lambda&1&2 \\1&\lambda&-1\\2&-1&\lambda \end{vmatrix}=0$

$\Rightarrow\lambda(\lambda^2-1)-1(\lambda+2)+2(-1-2\lambda)=0$

$\Rightarrow\lambda^3-\lambda-\lambda-2-2-4\lambda=0$

$\Rightarrow\lambda^3-6\lambda-4=0$

$\Rightarrow(\lambda+2)(\lambda^2-2\lambda-2)=0$

$\Rightarrow\lambda=-2$ or $\lambda=\frac{2\pm\sqrt{12}}{2}$

$\Rightarrow\lambda=-2$ or $\lambda=1\pm\sqrt{3}$

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 "M.C.Q (1 Marks)" from VECTOR ALGEBRA→VECTOR ALGEBRA — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Given that for each $a \in(0,1)$

$\lim _{n \rightarrow 0^{+}} \int_n^{1-n} t^{-3}(1-t)^{a-1} d t$

exists. Let this limit be $g(a)$. In addition, it is given that the function $g(a)$ is differentiable on $(0,1)$.

$1.$ The value of $g\left(\frac{1}{2}\right)$ is

$(A)$ $\pi$ $(B)$ $2 \pi$ $(C)$ $\frac{\pi}{2}$ $(D)$ $\frac{\pi}{4}$

$2.$ The value of $g ^{\prime}\left(\frac{1}{2}\right)$ is

$(A)$ $\frac{\pi}{2}$ $(B)$ $\pi$ $(C)$ $-\frac{\pi}{2}$ $(D)$ $0$

Give the answer question $1$ and $2.$

→
In a binomial distribution, the probability of getting success is $\frac{1}{4}$ and standard deviation is $3$. Then, its mean is:
→
The minimum value of $\frac{\text{x}}{\log_{\text{e}}\text{x}}$ is .
→
If $f(x) = 4x^8,$ then :
→
The slope of tangent to the curve $x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is:
→
If $\int \sin ^{-1}\left(\sqrt{\frac{x}{1+x}}\right) d x=A(x) \tan ^{-1}(\sqrt{x})+B(x)+C$ where $C$ is a constant of integration, then the ordered pair $( A ( x ), B ( x ))$ can be
→
The equation of the curve passing through the origin and satisfying the differential equation $\left( {1 + {x^2}} \right)\,\frac{{dy}}{{dx}} + 2xy = 4{x^2}$ is
→
If $|\vec{a}|=3,|\vec{b}|=4$ and $|\vec{a}+\vec{b}|=5$, then $|\vec{a}-\vec{b}|=$
→
$f(x) = \left\{ \begin{gathered}
  x,\,\,\,\,\,\,\,\,\,x\,{\text{is rational}} \hfill \\
  1 - x,\,\,x{\text{ is irrational}} \hfill \\ 
\end{gathered}  \right.$ then at $x = \frac{1}{{2,}}$ $f(x)$ is
→
Let $\times$ be a binary operation on set of integers I, defined by $a \times b = a + b - 3,$ then find the value of $3 \times 4.$
→