MCQ
If the vector $\vec a=i-j+2k ,\vec b=2i+4j+k $ and $\vec c=$ $\alpha i+j+\beta k$  are mutually orthogonal then $(\alpha ,\beta ) =$
  • A
    $(2,-3)$
  • B
    $(-2,3)$
  • C
    $(3,-2)$
  • $(-3,2)$

Answer

Correct option: D.
$(-3,2)$
d
$ \vec{a}=\hat{i}-\hat{j}+2 \hat{k}, $

$\vec{b}=2 \hat{i}+4 \hat{j}+4 \hat{k} ,$

$\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k} $

$\vec{a}$ and $\vec{c}$ are orthogonal $\Rightarrow \vec{a} \cdot \vec{c}=0$ giving $\lambda-1+2 \mu=0$

Also $\vec{b}$ and $\vec{c}$ are orthogonal $\Rightarrow 2 \lambda+4+4 \mu=0$

Solving the equation we get $\lambda=-3, \mu=2$

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

The value of $\left| {\,\begin{array}{*{20}{c}}1&{\cos (\beta - \alpha )}&{\cos (\gamma - \alpha )}\\{\cos (\alpha - \beta )}&1&{\cos (\gamma - \beta )}\\{\cos (\alpha - \gamma )}&{\cos (\beta - \gamma )}&1\end{array}} \right|$ is
$\sqrt i = $
Let $f:[0,2] \rightarrow R$ be the function defined by

$f ( x )=(3-\sin (2 \pi x )) \sin \left(\pi x -\frac{\pi}{4}\right)-\sin \left(3 \pi x +\frac{\pi}{4}\right)$

If $\alpha, \beta \in[0,2]$ are such that $\{x \in[0,2]: f(x) \geq 0\}=[\alpha, \beta]$, then the value of $\beta-\alpha$ is. . . . . . . . . 

Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1$. If $u$ and $v$ satisfy the equations
$
\begin{array}{l}
\alpha u+\beta v=18, \\
\gamma u+\delta v=20,
\end{array}
$
then $u+v$ equals :
$A$ and $B$ are two independent events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$. Then $P$ (neither $A$ nor $B$) is equal to
Let $P Q R$ be an acute-angled triangle in which $P Q < Q R$. From the vertex $Q$ draw the altitude $Q Q_1$, the angle bisector $Q Q_2$ and the median $Q Q_3$, with $Q_1, Q_2, Q_3$ lying on $P R$. Then,
Let : $\overrightarrow{ a }=\hat{ i }+2 \hat{ j }+3 \hat{ k }, \overrightarrow{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be there vectors. If $\vec{r}$ is a vector such that, $\overrightarrow{ r } \times \overrightarrow{ b }=\overrightarrow{ c } \times \overrightarrow{ b }$ and $\overrightarrow{ r } \cdot \overrightarrow{ a }=0$. Then $25|\overrightarrow{ r }|^2$ is equal to
Let $a, b$ and $c$ be distinct positive numbers. If the vectors $a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$ are co-planar, then $\mathrm{c}$ is equal to:
On the ellipse $4{x^2} + 9{y^2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are
$\int_{}^{} {\frac{{x - 2}}{{{x^2} - 4x + 3}}dx = } $