MCQ
Let $\vec a\, = \,\hat i\, + \,2\hat j\, + 4\hat k\,,\,\vec b\, = \,\hat i\, + \,\lambda \hat j\, + 4\hat k$ and $\vec c\, = \,2\hat i\, + \,4\hat j\, + ({\lambda ^2} - 1)\hat k$ be coplanar vectors. Then the non -zero vector $\vec a\times \vec c$ is
  • A
    $ - 10\,\,\hat i\, - 5\,j$
  • B
    $ - 14\,\,\hat i\, - 5\,j$
  • C
    $ - 14\,\,\hat i\, + 5\,j$
  • $ - 10\,\,\hat i\, + 5\,j$

Answer

Correct option: D.
$ - 10\,\,\hat i\, + 5\,j$
d
$\left[\begin{array}{lll}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{array}\right]=0$

$\left|\begin{array}{lll}{1} & {2} & {4} \\ {1} & {\lambda} & {4} \\ {2} & {4} & {\lambda^{2}-1}\end{array}\right|_{R_{3} \rightarrow R_{3}-2 R_{1}}=0$

$\left|\begin{array}{ccc}{1} & {2} & {4} \\ {1} & {\lambda} & {4} \\ {0} & {0} & {\lambda^{2}-9}\end{array}\right|=0$

$\left(\lambda^{2}-9\right)(\lambda-2)=0$

$\Rightarrow \lambda=2 \quad$ OR $\quad \lambda^{2}=5$

$\vec{a} \times \vec{c}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {1} & {2} & {4} \\ {2} & {4} & {\lambda^{2}-1}\end{array}\right|$

$=\hat{i}\left(2 \lambda^{2}-2-16\right)-\hat{j}\left(\lambda^{2}-1-8\right)$

$\left.=\left(\lambda^{2}-9\right)(2 \hat{i}-\hat{j})=-5(2 \hat{i}-\hat{j}) \quad \text { (put } \lambda=2\right)$

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

Similar questions

If the variable line $y = kx + 2h$ is tangent to an ellipse $2x^2 + 3y^2 = 6$ , then locus of $P(h, k)$ is a conic $C$ whose eccentricity equals
If $f(x) = \left[ \begin{gathered}
  {e^x} + a\,\,\,for\,\,\,x\, < \,0 \hfill \\
  x - 3\,\,\,\,\,for\,\,\,x\, \geqslant \,0 \hfill \\ 
\end{gathered}  \right.\,,\,$ is differentiable at $x = 0$ , then $'a'$ equals to 
If the function $f\,:\,R - \,\{ 1, - 1\}  \to A$ defined by $f\,(x)\, = \frac{{{x^2}}}{{1 - {x^2}}},$ is surjective, then $A$ is equal to
$\sqrt {[10 - \sqrt {(24)} - \sqrt {(40)} + \sqrt {(60)} ]} = $
If a curve passes through the point $\left( {2\,,\,\frac{7}{2}} \right)$ and has slope $\left( {1 - \frac{1}{{{x^2}}}} \right)$  at anypoint $(x, y)$ on it, then the ordinate of the point on the curve whose abscissa is $- 2$ is
The equation of the line passes through $(a,\;b)$and parallel to the line $\frac{x}{a} + \frac{y}{b} = 1,$is
Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group $B$, is equal to:
$\int_{1}^{6\pi}([sec^{-1}]+[cot^{-1}])dx$ is equal to       (where $[.]$ denotes greatest integer function)
The shortest distance between the lines $\frac{{x - 3}}{3} = \frac{{y - 8}}{{ - 1}} = \frac{{z - 3}}{1}$ and $\frac{{x + 3}}{{ - 3}} = \frac{{y + 7}}{2} = \frac{{z - 6}}{4}$ is 
If $P$ and $Q$ are the points of intersection of the circles ${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$ and ${x^2} + {y^2} + 2x + 2y - {p^2} = 0$ then there is a circle passing through $P, Q$ and $(1, 1)$ for: