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General

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Question

$P left parenthesis stack p with minus on top right parenthesis text end text$ and $Q left parenthesis stack q with minus on top right parenthesis$ are the position vectors of two fixed points and $R left parenthesis stack r with minus on top right parenthesis$ is the position vector of a variable point. If R moves such that $left parenthesis stack r with minus on top minus stack p with minus on top right parenthesis cross times left parenthesis stack r with minus on top minus stack q with minus on top right parenthesis$ =0 then the locus of R is

A plane containing the origin O and parallel to two non collinear vectors $\bar{O P}$ and $\bar{O Q}$
The surface of a sphere described on PQ as its diameter
A line passing through the points P and Q
A set of lines parallel to the line PQ

The correct answer is: A line passing through the points P and Q

Obviously (3). $R left parenthesis stack r with minus on top right parenthesis$ moves on PQ as $left parenthesis stack r with minus on top minus stack p with minus on top right parenthesis parallel to left parenthesis stack r with minus on top minus stack q with minus on top right parenthesis$

Related Questions to study

The functions $s i n to the power of 1 end exponent invisible function application x comma c o s to the power of 1 end exponent invisible function application x comma t a n to the power of 1 end exponent invisible function application x comma c o t to the power of 1 end exponent invisible function application x comma c o s e c to the power of 1 end exponent invisible function application x$ and $s e c to the power of 1 end exponent invisible function application x$ are called inverse circular or inverse trigonometric functions which are defined as follows
$s i n to the power of 1 end exponent invisible function application x text 1 end text x 1 pi divided by 2 s i n to the power of 1 end exponent invisible function application x pi divided by 2$
$c o s to the power of 1 end exponent cross times 1 cross times 10 c o s to the power of 1 end exponent cross times pi$
$t a n to the power of 1 end exponent invisible function application x cross times R pi divided by 2 less than t a n to the power of 1 end exponent invisible function application x less than pi divided by 2$
$c o s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 1 pi divided by 2 c o s e c to the power of 1 end exponent invisible function application x 0$
$s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 10 s e c to the power of 1 end exponent invisible function application x pi pi divided by 2$
$c o t to the power of 1 end exponent invisible function application x x R 0 less than c o t to the power of 1 end exponent invisible function application x less than pi$
Number of solutions of $t a n to the power of 1 end exponent invisible function application vertical line x vertical line c o s to the power of 1 end exponent invisible function application x equals 0$ is/are

The functions $s i n to the power of 1 end exponent invisible function application x comma c o s to the power of 1 end exponent invisible function application x comma t a n to the power of 1 end exponent invisible function application x comma c o t to the power of 1 end exponent invisible function application x comma c o s e c to the power of 1 end exponent invisible function application x$ and $s e c to the power of 1 end exponent invisible function application x$ are called inverse circular or inverse trigonometric functions which are defined as follows
$s i n to the power of 1 end exponent invisible function application x text 1 end text x 1 pi divided by 2 s i n to the power of 1 end exponent invisible function application x pi divided by 2$
$c o s to the power of 1 end exponent cross times 1 cross times 10 c o s to the power of 1 end exponent cross times pi$
$t a n to the power of 1 end exponent invisible function application x cross times R pi divided by 2 less than t a n to the power of 1 end exponent invisible function application x less than pi divided by 2$
$c o s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 1 pi divided by 2 c o s e c to the power of 1 end exponent invisible function application x 0$
$s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 10 s e c to the power of 1 end exponent invisible function application x pi pi divided by 2$
$c o t to the power of 1 end exponent invisible function application x x R 0 less than c o t to the power of 1 end exponent invisible function application x less than pi$
Number of solutions of $t a n to the power of 1 end exponent invisible function application vertical line x vertical line c o s to the power of 1 end exponent invisible function application x equals 0$ is/are

Given : $cos invisible function application xcos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x close parentheses cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x close parentheses equals fraction numerator 1 over denominator 4 end fraction comma x element of open square brackets 0 , 6 pi close square brackets$
Sum of all solutions is

Given : $cos invisible function application xcos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x close parentheses cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x close parentheses equals fraction numerator 1 over denominator 4 end fraction comma x element of open square brackets 0 , 6 pi close square brackets$
Sum of all solutions is

This section contains 3 multiple choice questions numbered 14 to 16 .Each question has 4 choices (A), (B),(c) , and (D), out of which ONE or MORE may be correct
If $fraction numerator 3 plus i 2 s i n invisible function application theta over denominator 1 minus i 2 s i n invisible function application theta end fraction$ is real , then

This section contains 3 multiple choice questions numbered 14 to 16 .Each question has 4 choices (A), (B),(c) , and (D), out of which ONE or MORE may be correct
If $fraction numerator 3 plus i 2 s i n invisible function application theta over denominator 1 minus i 2 s i n invisible function application theta end fraction$ is real , then

parallel

A charge q is placed at the center of the open end of cylindrical vessel. The flux of the electric field through surface of the vessel is

A charge q is placed at the center of the open end of cylindrical vessel. The flux of the electric field through surface of the vessel is

Physics-General

The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become

The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become

Physics-General

Figure shows four charges q₁, q₂, q₃ and q₄ fixed in space. Then the total flux of electric field through a closed surface S, due to all charges q₁, q₂, q₃ and q₄ is

Figure shows four charges q₁, q₂, q₃ and q₄ fixed in space. Then the total flux of electric field through a closed surface S, due to all charges q₁, q₂, q₃ and q₄ is

Physics-General

parallel

A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be

A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be

Physics-General

A Gaussian surface S enclosed two charges q₁ = q, q₂ = -q and q₃ = q. The net flux passing through S is

A Gaussian surface S enclosed two charges q₁ = q, q₂ = -q and q₃ = q. The net flux passing through S is

Physics-General

The ratio of $phi subscript E end subscript$ passing through the surface S₁ and S₂ is

The ratio of $phi subscript E end subscript$ passing through the surface S₁ and S₂ is

Physics-General

parallel

A point charge q is placed at the origin. The net flux of electric field (E) passing through the curved surface S is

A point charge q is placed at the origin. The net flux of electric field (E) passing through the curved surface S is

Physics-General

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in figure. The net electrostatic energy of the configuration is zero, if Q is equal to

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in figure. The net electrostatic energy of the configuration is zero, if Q is equal to

Physics-General

A charge Q is shifted from the origin to a point P(a, a). the work done by the electrical force is

A charge Q is shifted from the origin to a point P(a, a). the work done by the electrical force is

Physics-General

parallel

The charge -Q is shifted slowly from C to D, vertices of square of side l. The work done by the external agent is

The charge -Q is shifted slowly from C to D, vertices of square of side l. The work done by the external agent is

Physics-General

The field pattern is shown due to two point charges 1 and 2. Then q₁/q₂ =

The field pattern is shown due to two point charges 1 and 2. Then q₁/q₂ =

Physics-General

Statement : 1 The value of k for which the equation $x to the power of 2 end exponent plus k x minus fraction numerator pi over denominator 2 end fraction plus s i n to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses plus c o s to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses equals 0$ Has a real solution is -3.
Statement : 2 $s i n to the power of negative 1 end exponent invisible function application x plus c o s to the power of negative 1 end exponent invisible function application x equals fraction numerator pi over denominator 2 end fraction comma text for all end text x element of R$

Statement : 1 The value of k for which the equation $x to the power of 2 end exponent plus k x minus fraction numerator pi over denominator 2 end fraction plus s i n to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses plus c o s to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses equals 0$ Has a real solution is -3.
Statement : 2 $s i n to the power of negative 1 end exponent invisible function application x plus c o s to the power of negative 1 end exponent invisible function application x equals fraction numerator pi over denominator 2 end fraction comma text for all end text x element of R$

parallel

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