Download App

Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
Find the value of $\lambda$ so that the vectors $2 \hat{i}-4 \hat{j}+\hat{k}$ and $4 \hat{i}-8 \hat{j}+\lambda \hat{k}$ are perpendicular.
  • A
    20
  • ✓
    -40
  • C
    40
  • D
    -20
✓

Answer

Correct option: B.
-40
(b): The given vectors are perpendicular if their dot product vanishes, i.e.,
$
\begin{aligned}
& (2 \hat{i}-4 \hat{j}+\hat{k}) \cdot(4 \hat{i}-8 \hat{j}+\lambda \hat{k})=0 \\
\Rightarrow & 8+32+\lambda=0 \Rightarrow \lambda=-40
\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 "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

The distance travelled $s$ (in centi metre) by a particle in $ t $ seconds is given by, $s = {t^3} + 2{t^2} + t.$ The speed of the particle after $1 $ second will be ......... $cm/sec$
→
$\int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 3}}{{{x^2} + 4|x| + 3}}dx} $ is equal to -
→
The total number of matrices formed with the help of 6 different numbers are:
→
If $\text{f(x)}=\begin{cases}\text{ax}^2+\text{b},&0\leq\text{x} < 1\\4,&\text{x}=1\\\text{x}+3,&1<\text{x}\leq2\end{cases}$ then the value of $(a, b)$ for which $f(x)$ cannot be continuous at $x = 1,$ is :
→
The value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big),$ is:
→
Let $f(x) = \left[ {\begin{array}{*{20}{c}}
{g\,(x)\,\,.\,\,\cos \,{\textstyle{1 \over x}}}&{if\,\,\,\,x\,\, \ne \,\,0}\\
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}&{if\,\,\,\,x\,\, = \,\,0}
\end{array}} \right. $where $g(x)$ is an even function differentiable at $x = 0$, passing through the origin . Then $f ' (0)$ :
→
If $\text{y}=2\sin\text{x}+\sin2\text{x}$ for $0\leq \text{x}\leq 2\pi,$ then the area enclosed by the curve and $x-$ axis is :
→
Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $2 \hat{i}-3 \hat{j}+6 \hat{k}$.
→
Matrices $\mathrm{A}$ and $\mathrm{B}$ will be inverse of each other only if
→
The area of the region bounded by the parabola $y=x^2+1$ and the straight line $x + y = 3$ is given by:
→