Which of the following is an orthogonal matrix - Vidyadip

MCQ

Which of the following is an orthogonal matrix

✓
$\left[ {\begin{array}{*{20}{c}}{6/7}&{2/7}&{ - 3/7}\\{2/7}&{3/7}&{6/7}\\{3/7}&{ - 6/7}&{2/7}\end{array}} \right]$
B
$\left[ {\begin{array}{*{20}{c}}{6/7}&{2/7}&{3/7}\\{2/7}&{ - 3/7}&{6/7}\\{3/7}&{6/7}&{ - 2/7}\end{array}} \right]$
C
$\left[ {\begin{array}{*{20}{c}}{ - 6/7}&{ - 2/7}&{ - 3/7}\\{2/7}&{3/7}&{6/7}\\{ - 3/7}&{6/7}&{2/7}\end{array}} \right]$
D
$\left[ {\begin{array}{*{20}{c}}{6/7}&{ - 2/7}&{3/7}\\{2/7}&{2/7}&{ - 3/7}\\{ - 6/7}&{2/7}&{3/7}\end{array}} \right]$

✓

Answer

Correct option: A.

$\left[ {\begin{array}{*{20}{c}}{6/7}&{2/7}&{ - 3/7}\\{2/7}&{3/7}&{6/7}\\{3/7}&{ - 6/7}&{2/7}\end{array}} \right]$

a
Matrix $\left[ {\,\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}\,} \right]$ is orthogonal if

$\sum {a_i^2} \, = \,\sum {b_i^2} \, = \,\sum {c_i^2} \, = \,1$; $\sum {a_i\,b_i} = \sum {b_i\, c_i} = \sum {c_i\,a_i} = 0$

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 3 and 4 . determinant and metrices→3 and 4 . determinant and metrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $y=y(x)$ be the solution of the differential equation $\left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0$, $y(0)=1$. Then $y\left(\frac{\pi}{4}\right)$ is equal to :

Consider the family of all circles whose centers lie on the straight line $y =x$. If this family of circles is represented by the differential equation $P y^{\prime \prime}+Q y^{\prime}+1=0$, where $P, Q$ are functions of $x, y$ and $y^{\prime}$ (here $y^{\prime}=\frac{d y}{d x}, y ^{\prime \prime}=\frac{d^2 y}{d x^2}$ ), then which of the following statements is (are) true ?

$(A)$ $P=y+x$

$(B)$ $P=y-x$

$(C)$ $P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^2$

$(D)$ $P-Q=x+y-y^{\prime}-\left(y^{\prime}\right)^2$

Let a function $f(x)$ be defined as

$\begin{gathered}
f\left( x \right) = \left[ \begin{gathered}
{\cos ^{ - 1}}\left( \mu \right) + {x^2},0 < x < 1 \hfill \\
4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x \geqslant 1 \hfill \\
\end{gathered} \right.,f\left( x \right) \hfill \\
\hfill \\ \end{gathered}$ can have a local minimum at $x =$ $1$, if the value of $\mu$ lies in the interval

A particle moves in a straight line in such a way that its velocity at any point is given by ${v^2} = 2 - 3x$, where $x$ is measured from a fixed point. The acceleration is

Let $f (x) =$ $\mathop {Lim}\limits_{h \to 0} \,\,\frac{1}{h}\int\limits_x^{x + h} {\frac{{dt}}{{t + \sqrt {1 + {t^2}} }}} $, then $\mathop {Lim}\limits_{x \to \, - \infty } \,\,x\,\cdot\,{\rm{f}}(x)$ is

The total number of matrices formed with the help of 6 different numbers are:

6
6!
2(6!)
4(6!)

Considering only the principal values, if $\tan ({\cos ^{ - 1}}x)$ $ = \sin \left[ {{{\cot }^{ - 1}}\left( {\frac{1}{2}} \right)} \right]$, then $x$ is equal to

If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two collinear vectors, then which of the follwoing are incorrect?

$\vec{\text{b}}=\lambda\vec{\text{a}}$ for some scalar $\lambda$
$\vec{\text{a}}=\pm\vec{\text{b}}$
The respective components of $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are proportional.
Both the vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ have the same direction but different magnitudes.

Let for $i\, = 1, 2, 3, p_i(x)$ be a polynomial of degree $2$ in $x, p'_i(x)$ and $p"_i(x)$ be the first and second order derivatives of $p_i(x)$ respectively. Let, $A\left( x \right)=\left[ \begin{matrix}
{{p}_{1}}\left( x \right) & p_{1}^{'}\left( x \right) & p_{1}^{''}\left( x \right) \\
{{p}_{2}}\left( x \right) & p_{2}^{'}\left( x \right) & p_{2}^{''}\left( x \right) \\
{{p}_{3}}\left( x \right) & p_{3}^{'}\left( x \right) & p_{3}^{''}\left( x \right) \\
\end{matrix} \right]$ and $B(x)\,= [A(x)]^T$ $A(x)$. Then determinant of $B(x)$

If ${\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right|^2} = \left| {\,\begin{array}{*{20}{c}}3&2\\1&x\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{c}}x&3\\{ - 2}&1\end{array}\,} \right|$, then $ x =$