Download App

10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
MCQ
Let $\lambda \in R , \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}$ If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$, then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to
  • ✓
    $140$
  • B
    $132$
  • C
    $144$
  • D
    $136$
✓

Answer

Correct option: A.
$140$
a
$\overrightarrow{ a }=\lambda \hat{ i }+2 \hat{ j }-3 \hat{ k }$

$\overrightarrow{ b }=\hat{ i }-\lambda \hat{ j }+2 \hat{ k }$ $\Rightarrow(\overrightarrow{ b }-\overrightarrow{ a }) \times((\overrightarrow{ a }+\overrightarrow{ b }) \times(\overrightarrow{ a } \times \overrightarrow{ b }))=8 \hat{ i }-40 \hat{ j }-24 \hat{ k }$

$\Rightarrow((\overrightarrow{ a }-\overrightarrow{ b }) \cdot(\overrightarrow{ a }+\overrightarrow{ b }))(\overrightarrow{ a } \times \overrightarrow{ b })=8 \hat{ i }-40 j -24 \hat{ k }$

$\Rightarrow 8(\overrightarrow{ a } \times \overrightarrow{ b })=8 \hat{ i }-40 \hat{ j }-24 \hat{ k }$

$\text { Now, } \overrightarrow{ a } \times \overrightarrow{ b }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2\end{array}\right|$

$=(4-3 \lambda) \hat{ i }-(2 \lambda+3) \hat{ j }+\left(-\lambda^2-2\right) \hat{ k }$

$\Rightarrow \lambda=1$

$\therefore \overrightarrow{ a }=\hat{ i }+2 \hat{ j }-3 \hat{ k }$

$\overrightarrow{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$

$\Rightarrow \lambda=1$

$\therefore \overrightarrow{ a }=\hat{ i }+2 \hat{ j }-3 \hat{ k }$

$\overrightarrow{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$

$\Rightarrow \overrightarrow{ a }+\overrightarrow{ b }=2 \hat{ i }+\hat{ j }-\hat{ k }, \overrightarrow{ a }-\overrightarrow{ b }=3 \hat{ j }-5 \hat{ k }$ $\Rightarrow(\overrightarrow{ a }+\overrightarrow{ b }) \times(\overrightarrow{ a }-\overrightarrow{ b })=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -1 \\ 0 & 3 & -5\end{array}\right|=2$

$\hat{ i }+10 \hat{ j }+6 \hat{ k }$

$\therefore \text { required answer }=4+100+36=140$

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 10. vector algebra→10. 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 vector $\vec a = \left( {\alpha ,2,\beta } \right)\;,$ lies in the plane of the vector $\vec b = \left( {1,1,0} \right),$ and $\;\vec c = \left( {0,1,1} \right)$ and bisects the angle between $\vec b$  and $\vec c$ . Then which one of the following gives possible values $\alpha $ and $\beta $ ?
→
$\int\sec^2\text{x}.\text{cosec}^2\text{xdx}=$

  1. $\tan\text{x}-\cot\text{x+c}$

  2. $\tan\text{x}+\cot\text{x+c}$

  3. $-\tan\text{x}+\cot\text{x+c}$

  4. $\sec\text{x}\tan\text{x+c}$

→
A function $f$ satisfying $f ‘ \,(sin\, x)$ $= cos^2 x$ for all $x$ and $f(1) = 1$ is :
→
$\cos ^{-1}\left(\frac{-1}{2}\right)+2 \sin ^{-1}\left(\frac{-1}{2}\right)$ is equal to
→
$\int_{}^{} {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}\;dx} $ equals
→
If $a > 0$and discriminant of $a{x^2} + 2bx + c$is negative, then $\left| {\,\begin{array}{*{20}{c}}a&b&{ax + b}\\b&c&{bx + c}\\{ax + b}&{bx + c}&0\end{array}\,} \right|$ is
→
In Graphical solution the feasible solution is any solution to a LPP which satisfies.
  1. Only objective function.
  2. Non - negativity restriction.
  3. Only constraint.
  4. All the three
→
The differential equation $\cot y\,\,dx = x\,\,dy$ has a solution of the form
→
Area common to the curve $y =\sqrt {9\, - \,{x^2}} $  $\& \,\,x^2 + y^2 = 6$ $x$ is :
→
The number of binary operation that can be defined on a set of 2 elements is:
  1. 8
  2. 4
  3. 16
  4. 64
→