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Maths-

General

Easy

Question

Assertion (A) : If $stack a with minus on top equals 2 i plus stack k with minus on top comma stack b with minus on top equals 3 j plus 4 k text end text$ and $stack c with minus on top equals 8 i minus 3 j$ are coplanar then $stack c with minus on top equals 4 stack a with minus on top minus stack b with minus on top$
Reason (R) : A set of vectors a₁, a₂, a₃.a_n is said to be linearly dependent if for some relation of the form $l subscript 1 end subscript a subscript 1 end subscript plus l subscript 2 end subscript stack a with minus on top subscript 2 end subscript plus l horizontal ellipsis horizontal ellipsis. plus 1 subscript n end subscript stack a with minus on top subscript n end subscript equals stack 0 with bar on top$ implies at least one of the scalars l_i (i = 1,2.n) is not zero

Both A and R are true and R is the correct explanation of A
Both A and R are true but R is not the correct explanation of A
A is true, R is false
A is false, R is true

The correct answer is: Both A and R are true and R is the correct explanation of A

Conceptual

Related Questions to study

Let $stack a with rightwards arrow on top equals stack i with hat on top plus stack j with hat on top plus 2 stack k with hat on top text end text$ and $stack b with rightwards arrow on top equals 2 stack i with hat on top minus stack j with hat on top minus stack k with hat on top$ . Then the point of intersection of the
lines $stack r with rightwards arrow on top cross times stack a with rightwards arrow on top equals stack b with rightwards arrow on top cross times stack a with rightwards arrow on top$ and $stack r with rightwards arrow on top cross times stack b with rightwards arrow on top equals stack a with rightwards arrow on top cross times stack b with rightwards arrow on top$ is

Let $stack a with rightwards arrow on top equals stack i with hat on top plus stack j with hat on top plus 2 stack k with hat on top text end text$ and $stack b with rightwards arrow on top equals 2 stack i with hat on top minus stack j with hat on top minus stack k with hat on top$ . Then the point of intersection of the
lines $stack r with rightwards arrow on top cross times stack a with rightwards arrow on top equals stack b with rightwards arrow on top cross times stack a with rightwards arrow on top$ and $stack r with rightwards arrow on top cross times stack b with rightwards arrow on top equals stack a with rightwards arrow on top cross times stack b with rightwards arrow on top$ is

The length of the projection of the line segment joining the points A (1,2,5) and B (4,7,3) on the line whose direction ratios are 6,2,3 is

The length of the projection of the line segment joining the points A (1,2,5) and B (4,7,3) on the line whose direction ratios are 6,2,3 is

If the four plane focus of a tetrahedron are represented by the equation lx + my = 0, my + nz = 0, nz + px = 0, and lx + my + nz = p , then the volume of the tetrahedron are

If the four plane focus of a tetrahedron are represented by the equation lx + my = 0, my + nz = 0, nz + px = 0, and lx + my + nz = p , then the volume of the tetrahedron are

parallel

If $stack a with minus on top$ and $stack b with minus on top$ are two unit vectors and $ϕ$ is the angle between them, then $fraction numerator 1 over denominator 2 end fraction vertical line stack a with minus on top minus stack b with minus on top vertical line$ is equal to

If $stack a with minus on top$ and $stack b with minus on top$ are two unit vectors and $ϕ$ is the angle between them, then $fraction numerator 1 over denominator 2 end fraction vertical line stack a with minus on top minus stack b with minus on top vertical line$ is equal to

Statement-1:The area of the triangle formed by the points A(20,22) ;B(21,24) and C(22,23) is the same as the area of the triangle formed by the point P(0,0) ;Q(1,2) and R(2,1) . Because
Statement-2: The area of the triangle is invariant w.r.t translation of the coordinate axes.

Statement-1:The area of the triangle formed by the points A(20,22) ;B(21,24) and C(22,23) is the same as the area of the triangle formed by the point P(0,0) ;Q(1,2) and R(2,1) . Because
Statement-2: The area of the triangle is invariant w.r.t translation of the coordinate axes.

$stack a with minus on top equals stack i with minus on top plus stack j with minus on top plus stack k with minus on top comma stack b with minus on top equals 2 stack i with minus on top plus 3 stack j with minus on top plus 5 stack k with minus on top$ and $stack c with minus on top$ is a unit vector. The maximum value of the scalar triple product $left square bracket stack a b c with bar on top right square bracket$ is

$stack a with minus on top equals stack i with minus on top plus stack j with minus on top plus stack k with minus on top comma stack b with minus on top equals 2 stack i with minus on top plus 3 stack j with minus on top plus 5 stack k with minus on top$ and $stack c with minus on top$ is a unit vector. The maximum value of the scalar triple product $left square bracket stack a b c with bar on top right square bracket$ is

parallel

In metals the time of relaxation of electrons

In metals the time of relaxation of electrons

Physics-General

$P left parenthesis stack p with minus on top right parenthesis text end text$ 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

$P left parenthesis stack p with minus on top right parenthesis text end text$ 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

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

parallel

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

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

parallel

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

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

parallel

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