Maths-
General
Easy

Question

A vector has components P and 1 with respect to a rectangular Cartesian system. If the axes are rotated through an angle  about the origin in the anticlockwise sense.
Statement-1 : If the vector has component P + 2 and 1 with respect to the new system then P = –1
Statement-2 : Magnitude of vector original and new system remains same.

  1. If both Statement-1 and Statement-2 are true and the Statement-2 is correct explanation of the Statement-1.  
  2. If both Statement-1 and Statement-2 are true but Statement-2 is not correct explanation of the Statement-1.  
  3. If Statement-1 is true but the Statement-2 is false.  
  4. If Statement-1 is false but Statement-2 is true  

The correct answer is: If Statement-1 is false but Statement-2 is true

Related Questions to study

General
Maths-

Statement I : If three points P, Q, R have position vectorsstack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on toprespectively and2 stack a with rightwards arrow on top plus 3 stack b with rightwards arrow on top – 5 stack c with rightwards arrow on top equals 0, then the points P, Q, R must be collinear.
Statement II : If for three points A, B, C, ; Error converting from MathML to accessible text., then the points A, B, C must be collinear.

Statement I : If three points P, Q, R have position vectorsstack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on toprespectively and2 stack a with rightwards arrow on top plus 3 stack b with rightwards arrow on top – 5 stack c with rightwards arrow on top equals 0, then the points P, Q, R must be collinear.
Statement II : If for three points A, B, C, ; Error converting from MathML to accessible text., then the points A, B, C must be collinear.

Maths-General
General
Maths-

A vector has components p and 1 with respect to a rectangular Cartesian system. If the axes are rotated through an angle  about the origin in the anticlockwise sense.
Statement I : If the vector has component p + 2 and 1 with respect to the new system then p = –1
Statement II : Magnitude of vector with original
and new system remains same

A vector has components p and 1 with respect to a rectangular Cartesian system. If the axes are rotated through an angle  about the origin in the anticlockwise sense.
Statement I : If the vector has component p + 2 and 1 with respect to the new system then p = –1
Statement II : Magnitude of vector with original
and new system remains same
Maths-General

General
Maths-

Assertion: The point of intersection of the lines
Reason : Skew lines do not intersect.

Assertion: The point of intersection of the lines
Reason : Skew lines do not intersect.

Maths-General
parallel
General
Maths-

Let P, Q and R are points on sides AB, AC and AD of the parallelogram ABCD such that Error converting from MathML to accessible text.and Error converting from MathML to accessible text., where k1, k2 and k3 are non-zero positive scalars
Assertion(A) : k1, 2k2 and k3 are in harmonic progression if P, Q and R are collinear
Reason(R) :

Let P, Q and R are points on sides AB, AC and AD of the parallelogram ABCD such that Error converting from MathML to accessible text.and Error converting from MathML to accessible text., where k1, k2 and k3 are non-zero positive scalars
Assertion(A) : k1, 2k2 and k3 are in harmonic progression if P, Q and R are collinear
Reason(R) :
Maths-General

General
Maths-

Assertion(A): If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top then equation stack r with rightwards arrow on top cross times left parenthesis 2 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top right parenthesis equals 3 stack i with hat on top plus stack k with hat on top represent a straight line.
Reason(R): If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top, then equation stack r with rightwards arrow on top cross times left parenthesis stack i with hat on top plus 2 stack j with hat on top minus 3 stack k with hat on top right parenthesis equals 2 stack i with hat on top minus stack j with hat on top represent a straight line

Assertion(A): If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top then equation stack r with rightwards arrow on top cross times left parenthesis 2 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top right parenthesis equals 3 stack i with hat on top plus stack k with hat on top represent a straight line.
Reason(R): If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top, then equation stack r with rightwards arrow on top cross times left parenthesis stack i with hat on top plus 2 stack j with hat on top minus 3 stack k with hat on top right parenthesis equals 2 stack i with hat on top minus stack j with hat on top represent a straight line

Maths-General
General
Maths-

Assertion(A): Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis and C left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top equals 2 stack i with hat on top plus stack j with hat on top plus stack k with hat on top comma stack b with rightwards arrow on top equals 3 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top and stack c with rightwards arrow on top equals – stack i with hat on top plus 7 stack j with hat on top minus 5 stack k with hat on top then OABC is a tetrahedron.
Reason(R): Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis andC left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top are non-coplanar, then OABC is a tetrahedron, where O is the origin.

Assertion(A): Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis and C left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top equals 2 stack i with hat on top plus stack j with hat on top plus stack k with hat on top comma stack b with rightwards arrow on top equals 3 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top and stack c with rightwards arrow on top equals – stack i with hat on top plus 7 stack j with hat on top minus 5 stack k with hat on top then OABC is a tetrahedron.
Reason(R): Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis andC left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top are non-coplanar, then OABC is a tetrahedron, where O is the origin.

Maths-General
parallel
General
Maths-

Assertion: If stack a with ̄ on top× stack b with ̄ on top= stack c with ̄ on top× stack d with ̄ on top, and stack a with ̄ on top×stack c with ̄ on top= stack b with ̄ on top× stack d with ̄ on top then stack a with ̄ on topstack d with ̄ on top is perpendicular to stack b with ̄ on topstack c with ̄ on top.
Reason: If stack r with ̄ on topis perpendicular tostack q with ̄ on topthen stack r with ̄ on top.stack q with ̄ on top= 0

Assertion: If stack a with ̄ on top× stack b with ̄ on top= stack c with ̄ on top× stack d with ̄ on top, and stack a with ̄ on top×stack c with ̄ on top= stack b with ̄ on top× stack d with ̄ on top then stack a with ̄ on topstack d with ̄ on top is perpendicular to stack b with ̄ on topstack c with ̄ on top.
Reason: If stack r with ̄ on topis perpendicular tostack q with ̄ on topthen stack r with ̄ on top.stack q with ̄ on top= 0

Maths-General
General
Maths-

Assertion: Vectors – 2stack i with hat on top+stack j with hat on top + stack k with hat on top, stack i with hat on top–stack j with hat on top+stack k with hat on topand stack i with hat on top+ stack j with hat on top –2stack k with hat on top are coplanar for only two values of .
Reason: Three vectors stack a with rightwards arrow on top, stack b with rightwards arrow on top, stack c with rightwards arrow on topare coplanar if stack a with rightwards arrow on top.(stack b with rightwards arrow on top× stack c with rightwards arrow on top) = stack 0 with rightwards arrow on top.

Assertion: Vectors – 2stack i with hat on top+stack j with hat on top + stack k with hat on top, stack i with hat on top–stack j with hat on top+stack k with hat on topand stack i with hat on top+ stack j with hat on top –2stack k with hat on top are coplanar for only two values of .
Reason: Three vectors stack a with rightwards arrow on top, stack b with rightwards arrow on top, stack c with rightwards arrow on topare coplanar if stack a with rightwards arrow on top.(stack b with rightwards arrow on top× stack c with rightwards arrow on top) = stack 0 with rightwards arrow on top.

Maths-General
General
Maths-

Assertion (A): If vector stack a with rightwards arrow on top and stack b with rightwards arrow on topare linearly dependent, then vectors stack a with rightwards arrow on top, stack b with rightwards arrow on top, stack c with rightwards arrow on top must be dependent.
Reason (R): If vector stack a with rightwards arrow on top and stack b with rightwards arrow on top are linearly independent, then vectors stack a with rightwards arrow on top, stack b with rightwards arrow on top, stack c with rightwards arrow on top must be linearly independent, where vector stack c with rightwards arrow on top is non-zero.

Assertion (A): If vector stack a with rightwards arrow on top and stack b with rightwards arrow on topare linearly dependent, then vectors stack a with rightwards arrow on top, stack b with rightwards arrow on top, stack c with rightwards arrow on top must be dependent.
Reason (R): If vector stack a with rightwards arrow on top and stack b with rightwards arrow on top are linearly independent, then vectors stack a with rightwards arrow on top, stack b with rightwards arrow on top, stack c with rightwards arrow on top must be linearly independent, where vector stack c with rightwards arrow on top is non-zero.

Maths-General
parallel
General
Maths-

Assertion: If in a ABC ; Error converting from MathML to accessible text. = Error converting from MathML to accessible text.Error converting from MathML to accessible text. and Error converting from MathML to accessible text. = Error converting from MathML to accessible text. ; |Error converting from MathML to accessible text.|  |Error converting from MathML to accessible text.|, then the value of cos 2A + cos 2B + cos 2C is – 1.
Reason: If in ABC, C = 90º, then cos 2A + cos 2B + cos 2C = – 1.

Assertion: If in a ABC ; Error converting from MathML to accessible text. = Error converting from MathML to accessible text.Error converting from MathML to accessible text. and Error converting from MathML to accessible text. = Error converting from MathML to accessible text. ; |Error converting from MathML to accessible text.|  |Error converting from MathML to accessible text.|, then the value of cos 2A + cos 2B + cos 2C is – 1.
Reason: If in ABC, C = 90º, then cos 2A + cos 2B + cos 2C = – 1.
Maths-General

General
Maths-

If stack a with ̄ on top comma stack b with ̄ on top comma stack c with ̄ on top are noncoplanar vectors and stack r with rightwards arrow on top equals left parenthesis stack a with ̄ on top cross times stack b with ̄ on top right parenthesis cross times left parenthesis stack a with ̄ on top cross times stack c with ̄ on top right parenthesis.
Assertion: stack r with ̄ on topand stack a with ̄ on top are linearly dependent
Reason: stack r with rightwards arrow on topis r to each of three stack a with ̄ on top comma stack b with ̄ on top comma & stack c with ̄ on top.

If stack a with ̄ on top comma stack b with ̄ on top comma stack c with ̄ on top are noncoplanar vectors and stack r with rightwards arrow on top equals left parenthesis stack a with ̄ on top cross times stack b with ̄ on top right parenthesis cross times left parenthesis stack a with ̄ on top cross times stack c with ̄ on top right parenthesis.
Assertion: stack r with ̄ on topand stack a with ̄ on top are linearly dependent
Reason: stack r with rightwards arrow on topis r to each of three stack a with ̄ on top comma stack b with ̄ on top comma & stack c with ̄ on top.

Maths-General
General
Maths-

Maths-General

parallel
General
Maths-

Assertion(A) : If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top then equation stack r with rightwards arrow on top cross times left parenthesis 2 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top right parenthesis equals 3 stack i with hat on top plus stack k with hat on top represent a straight line.
Reason(R) : If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top, then equation stack r with rightwards arrow on top cross times left parenthesis stack i with hat on top plus 2 stack j with hat on top minus 3 stack k with hat on top right parenthesis equals 2 stack i with hat on top minus stack j with hat on top represent a straight line

Assertion(A) : If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top then equation stack r with rightwards arrow on top cross times left parenthesis 2 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top right parenthesis equals 3 stack i with hat on top plus stack k with hat on top represent a straight line.
Reason(R) : If stack r with rightwards arrow on top equals x stack i with hat on top plus y stack j with hat on top plus z stack k with hat on top, then equation stack r with rightwards arrow on top cross times left parenthesis stack i with hat on top plus 2 stack j with hat on top minus 3 stack k with hat on top right parenthesis equals 2 stack i with hat on top minus stack j with hat on top represent a straight line

Maths-General
General
Maths-

Assertion(A) : Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis and C left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top equals 2 stack i with hat on top plus stack j with hat on top plus stack k with hat on top comma stack b with rightwards arrow on top equals 3 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top and stack c with rightwards arrow on top equals – stack i with hat on top plus 7 stack j with hat on top minus 5 stack k with hat on top then OABC is a tetrahedron.
Reason(R) : Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis andC left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top are non-coplanar, then OABC is a tetrahedron, where O is the origin.

Assertion(A) : Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis and C left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top equals 2 stack i with hat on top plus stack j with hat on top plus stack k with hat on top comma stack b with rightwards arrow on top equals 3 stack i with hat on top minus stack j with hat on top plus 3 stack k with hat on top and stack c with rightwards arrow on top equals – stack i with hat on top plus 7 stack j with hat on top minus 5 stack k with hat on top then OABC is a tetrahedron.
Reason(R) : Let A left parenthesis stack a with rightwards arrow on top right parenthesis comma B left parenthesis stack b with rightwards arrow on top right parenthesis andC left parenthesis stack c with rightwards arrow on top right parenthesis be three points such that stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top are non-coplanar, then OABC is a tetrahedron, where O is the origin.

Maths-General
General
Maths-

Assertion : If stack a with ̄ on top× stack b with ̄ on top= stack c with ̄ on top× stack d with ̄ on top, and stack a with ̄ on top×stack c with ̄ on top= stack b with ̄ on top× stack d with ̄ on top then stack a with ̄ on topstack d with ̄ on top is perpendicular to stack b with ̄ on topstack c with ̄ on top.
Reason : If stack r with ̄ on topis perpendicular tostack q with ̄ on topthen stack r with ̄ on top.stack q with ̄ on top= 0

Assertion : If stack a with ̄ on top× stack b with ̄ on top= stack c with ̄ on top× stack d with ̄ on top, and stack a with ̄ on top×stack c with ̄ on top= stack b with ̄ on top× stack d with ̄ on top then stack a with ̄ on topstack d with ̄ on top is perpendicular to stack b with ̄ on topstack c with ̄ on top.
Reason : If stack r with ̄ on topis perpendicular tostack q with ̄ on topthen stack r with ̄ on top.stack q with ̄ on top= 0

Maths-General
parallel

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