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Maths-

General

Easy

Question

The equation of a plane and condition of two planes being parallel or perpendicular.
A) Every equation of first degree in x, y, z, i. e., Ax + By + Cz + D = 0 represents a plane. The coefficients of x, y, z are the direction ratios of the normal to the plane.
B) Angle between two planes is equal to the angle between the normals to the planes.
\ cos q = $fraction numerator A subscript 1 end subscript A subscript 2 end subscript plus B subscript 1 end subscript B subscript 2 end subscript plus C subscript 1 end subscript C subscript 2 end subscript over denominator square root of not stretchy sum A subscript 1 end subscript superscript 2 end superscript end root. square root of not stretchy sum A subscript 2 end subscript superscript 2 end superscript end root end fraction$
Planes are perpendicular if å A₁ A₂ = 0 and parallel if $fraction numerator A subscript 1 end subscript over denominator A subscript 2 end subscript end fraction equals fraction numerator B subscript 1 end subscript over denominator B subscript 2 end subscript end fraction equals fraction numerator C subscript 1 end subscript over denominator C subscript 2 end subscript end fraction.$
C) Planes parallel to co – ordinate planes are x = l, y = l or z = l.
Planes perpendicular to co – ordinate planes x = 0, y = 0, z = 0 are
by + cz + d = 0 (x missing), ax + cz + d = 0
(y missing), ax + by + d = 0 (z missing) Find the equation of the plane through the intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is

7x – 2y + 3z + 81 = 0
23x + 14y – 9z + 48 = 0
51x + 15y – 50z + 173 = 0
None

The correct answer is: 51x + 15y – 50z + 173 = 0

Related Questions to study

$stack lim with x blank rightwards arrow fraction numerator pi over denominator 2 end fraction below invisible function application open parentheses fraction numerator 1 plus c o s x over denominator 1 minus c o s x end fraction close parentheses to the power of s e c x end exponent equals$

$stack lim with x blank rightwards arrow fraction numerator pi over denominator 2 end fraction below invisible function application open parentheses fraction numerator 1 plus c o s x over denominator 1 minus c o s x end fraction close parentheses to the power of s e c x end exponent equals$

The direction cosines of the line joining the points (4, 3, - 5) and (-2, 1, -8) are

For such questions, we should know formula to find direction cosines.

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For such questions, we should know formula to find direction cosines.

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For such questions, we have to be careful about points satisfying the equation.

The locus of x² + y² + z² = 0 is

For such questions, we have to be careful about points satisfying the equation.

parallel

$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator 1 over denominator square root of x to the power of 2 end exponent plus 4 x plus 7 end root minus x end fraction equals$

$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator 1 over denominator square root of x to the power of 2 end exponent plus 4 x plus 7 end root minus x end fraction equals$

$stack lim with x blank rightwards arrow 0 below invisible function application x c o s fraction numerator 1 over denominator x end fraction equals$

$stack lim with x blank rightwards arrow 0 below invisible function application x c o s fraction numerator 1 over denominator x end fraction equals$

Let f (x) be a quadratic expression which is positive for all real x. If g(x) = f (x) + f ¢ (x) + f ¢¢ (x), then for any real x

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parallel

$stack lim with x blank rightwards arrow 0 below invisible function application fraction numerator 1 minus c o s a x over denominator x open parentheses e to the power of b x end exponent minus 1 close parentheses end fraction equals$

$stack lim with x blank rightwards arrow 0 below invisible function application fraction numerator 1 minus c o s a x over denominator x open parentheses e to the power of b x end exponent minus 1 close parentheses end fraction equals$

$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator open square brackets x close square brackets over denominator x end fraction left parenthesis W h e r e blank open square brackets. close square brackets blank d e n o t e s blank g r e a t e s t blank i n t e g e r blank f u n c t i o n right parenthesis$ =

$stack lim with x blank rightwards arrow infinity below invisible function application fraction numerator open square brackets x close square brackets over denominator x end fraction left parenthesis W h e r e blank open square brackets. close square brackets blank d e n o t e s blank g r e a t e s t blank i n t e g e r blank f u n c t i o n right parenthesis$ =

$I f stack lim with x blank rightwards arrow 5 below invisible function application fraction numerator x to the power of k end exponent minus 5 to the power of k end exponent over denominator x minus 5 end fraction equals 500 comma blank t h e n blank t h e blank p o s i t i v e blank i n t e g r a l blank v a l u e blank o f blank k blank i s$

$I f stack lim with x blank rightwards arrow 5 below invisible function application fraction numerator x to the power of k end exponent minus 5 to the power of k end exponent over denominator x minus 5 end fraction equals 500 comma blank t h e n blank t h e blank p o s i t i v e blank i n t e g r a l blank v a l u e blank o f blank k blank i s$

parallel

$If blank f open parentheses 9 close parentheses equals 9 comma blank f to the power of 1 end exponent open parentheses 9 close parentheses equals 4 comma blank stack lim with x blank rightwards arrow 9 below invisible function application fraction numerator square root of f open parentheses x close parentheses end root minus 3 over denominator square root of x minus 3 end fraction equals$

$If blank f open parentheses 9 close parentheses equals 9 comma blank f to the power of 1 end exponent open parentheses 9 close parentheses equals 4 comma blank stack lim with x blank rightwards arrow 9 below invisible function application fraction numerator square root of f open parentheses x close parentheses end root minus 3 over denominator square root of x minus 3 end fraction equals$

$I f blank stack text Lim end text with text x end text rightwards arrow a plus below invisible function application text f(x) end text equals k blank a n d blank stack text Lim end text with text x end text rightwards arrow a minus below invisible function application text f(x) = end text text l then end text stack text Lim end text with text x end text rightwards arrow a below invisible function application text f(x) = end text text end text blank$

For such questions, we should know different the condition of limit to exist.

$I f blank stack text Lim end text with text x end text rightwards arrow a plus below invisible function application text f(x) end text equals k blank a n d blank stack text Lim end text with text x end text rightwards arrow a minus below invisible function application text f(x) = end text text l then end text stack text Lim end text with text x end text rightwards arrow a below invisible function application text f(x) = end text text end text blank$

For such questions, we should know different the condition of limit to exist.

$c o t to the power of negative 1 end exponent open parentheses square root of c o s alpha end root close parentheses minus t a n to the power of negative 1 end exponent open parentheses square root of c o s alpha end root close parentheses equals x blank greater or equal than 0 blank$ then sinx =

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parallel

If $t a n open parentheses s e c to the power of negative 1 end exponent fraction numerator text 1 end text over denominator text x end text end fraction close parentheses equals s i n open parentheses T a n to the power of negative 1 end exponent 2 close parentheses$ then x=

If $t a n open parentheses s e c to the power of negative 1 end exponent fraction numerator text 1 end text over denominator text x end text end fraction close parentheses equals s i n open parentheses T a n to the power of negative 1 end exponent 2 close parentheses$ then x=

If 2 $T a n to the power of negative 1 end exponent open parentheses t a n blank alpha blank t a n blank beta close parentheses blank$ = x then x =

If 2 $T a n to the power of negative 1 end exponent open parentheses t a n blank alpha blank t a n blank beta close parentheses blank$ = x then x =

I : $T a n to the power of negative 1 end exponent 2$ + $T a n to the power of negative 1 end exponent 3 equals fraction numerator text 3π end text over denominator text 4 end text end fraction$
II : $c o s open curly brackets C o s to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses plus S i n to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses close curly brackets$ = 0
Which of the above statements is correct?

I : $T a n to the power of negative 1 end exponent 2$ + $T a n to the power of negative 1 end exponent 3 equals fraction numerator text 3π end text over denominator text 4 end text end fraction$
II : $c o s open curly brackets C o s to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses plus S i n to the power of negative 1 end exponent open parentheses fraction numerator text -1 end text over denominator text 7 end text end fraction close parentheses close curly brackets$ = 0
Which of the above statements is correct?

parallel

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