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$text If end text stack a with rightwards arrow on top equals i plus j minus k comma stack b with rightwards arrow on top equals i minus j plus k comma stack c with rightwards arrow on top$ is a unit vector such that $stack c with rightwards arrow on top times stack a with rightwards arrow on top equals 0 comma left square bracket stack c with rightwards arrow on top stack a with rightwards arrow on top stack b with rightwards arrow on top right square bracket equals$ 0 then a unit vector $text end text stack d with rightwards arrow on top$ perpendicular to $text both end text stack a with rightwards arrow on top text and end text stack c with rightwards arrow on top text end text text is end text$

$\frac{1}{\sqrt{6}} (2 i - j + k)$
$\frac{1}{2} (j + k)$
$\frac{1}{\sqrt{2}} (i + j)$
$\frac{1}{\sqrt{2}} (i + k)$

The correct answer is: $fraction numerator 1 over denominator 2 end fraction left parenthesis j plus k right parenthesis$

Related Questions to study

The reflection of the point (2, –1, 3) in the plane 3x – 2y – z = 9 is :

The reflection of the point (2, –1, 3) in the plane 3x – 2y – z = 9 is :

A, B, C & D are four points in a plane with position vectors $stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top & stack d with rightwards arrow on top$ respectively such that $left parenthesis stack a with rightwards arrow on top minus stack d with rightwards arrow on top right parenthesis times left parenthesis stack b with rightwards arrow on top minus stack c with rightwards arrow on top right parenthesis equals left parenthesis stack b with rightwards arrow on top minus stack d with rightwards arrow on top right parenthesis times left parenthesis stack c with rightwards arrow on top minus stack a with rightwards arrow on top right parenthesis equals 0$ Then for the triangle ABC, D is its:

A, B, C & D are four points in a plane with position vectors $stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top & stack d with rightwards arrow on top$ respectively such that $left parenthesis stack a with rightwards arrow on top minus stack d with rightwards arrow on top right parenthesis times left parenthesis stack b with rightwards arrow on top minus stack c with rightwards arrow on top right parenthesis equals left parenthesis stack b with rightwards arrow on top minus stack d with rightwards arrow on top right parenthesis times left parenthesis stack c with rightwards arrow on top minus stack a with rightwards arrow on top right parenthesis equals 0$ Then for the triangle ABC, D is its:

Equation of the angle bisector of the angle between the lines $fraction numerator x minus 1 over denominator 1 end fraction equals fraction numerator y minus 2 over denominator 1 end fraction equals fraction numerator z minus 3 over denominator 1 end fraction$ $fraction numerator x minus 1 over denominator 1 end fraction equals fraction numerator y minus 2 over denominator 1 end fraction equals fraction numerator z minus 3 over denominator negative 1 end fraction text is : end text$

Equation of the angle bisector of the angle between the lines $fraction numerator x minus 1 over denominator 1 end fraction equals fraction numerator y minus 2 over denominator 1 end fraction equals fraction numerator z minus 3 over denominator 1 end fraction$ $fraction numerator x minus 1 over denominator 1 end fraction equals fraction numerator y minus 2 over denominator 1 end fraction equals fraction numerator z minus 3 over denominator negative 1 end fraction text is : end text$

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The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where A º (3, –2, 1) ; B º (3, 1, 5); C º (4, 0, 3) and D º (1, 0, 0) is -

The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where A º (3, –2, 1) ; B º (3, 1, 5); C º (4, 0, 3) and D º (1, 0, 0) is -

Four coplanar forces are applied at a point O. Each of them is equal to k and the angle between two consecutive forces equals 45º as shown in the figure. Then the resultant has the magnitude equal to :

Four coplanar forces are applied at a point O. Each of them is equal to k and the angle between two consecutive forces equals 45º as shown in the figure. Then the resultant has the magnitude equal to :

$text Let end text stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top text end text$ be vectors of length 3,4,5 respectively. $text Let end text stack a with rightwards arrow on top text end text$ be perpendicular to $text end text stack b with rightwards arrow on top plus stack c with rightwards arrow on top comma stack b with rightwards arrow on top text to end text stack c with rightwards arrow on top plus stack a with rightwards arrow on top text end text$ and $text c to end text stack a with rightwards arrow on top plus stack b with rightwards arrow on top text . Then end text vertical line stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack c with rightwards arrow on top vertical line$ is equal to :

$text Let end text stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top text end text$ be vectors of length 3,4,5 respectively. $text Let end text stack a with rightwards arrow on top text end text$ be perpendicular to $text end text stack b with rightwards arrow on top plus stack c with rightwards arrow on top comma stack b with rightwards arrow on top text to end text stack c with rightwards arrow on top plus stack a with rightwards arrow on top text end text$ and $text c to end text stack a with rightwards arrow on top plus stack b with rightwards arrow on top text . Then end text vertical line stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack c with rightwards arrow on top vertical line$ is equal to :

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$text If end text vertical line stack a with rightwards arrow on top vertical line equals 5 comma vertical line stack a with rightwards arrow on top minus stack b with rightwards arrow on top vertical line equals 8 text end text$ and $vertical line stack a with rightwards arrow on top plus stack b with rightwards arrow on top vertical line equals 10 text , then end text vertical line stack b with rightwards arrow on top vertical line$ is equal to :

$text If end text vertical line stack a with rightwards arrow on top vertical line equals 5 comma vertical line stack a with rightwards arrow on top minus stack b with rightwards arrow on top vertical line equals 8 text end text$ and $vertical line stack a with rightwards arrow on top plus stack b with rightwards arrow on top vertical line equals 10 text , then end text vertical line stack b with rightwards arrow on top vertical line$ is equal to :

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If ABCDEF is a regular hexagon and if $stack A B with bar on top plus stack A C with rightwards arrow on top plus stack A D with rightwards arrow on top plus stack A E with rightwards arrow on top plus stack A F with rightwards arrow on top equals lambda stack A D with bar on top comma text end text$ then $text end text lambda text is - end text$

If $text end text stack a with rightwards arrow on top plus stack b with rightwards arrow on top text end text$ is along the angle bisector of $stack a with rightwards arrow on top & stack b with rightwards arrow on top text end text$ then -

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parallel

If the solution of the differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator x c o s invisible function application y plus s i n invisible function application 2 y end fraction text end text text is end text text end text x equals c o s i n invisible function application y minus k left parenthesis 1 plus s i n invisible function application y right parenthesis comma text then end text k equals$

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The solution of the differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator x plus y over denominator x end fraction$ $fraction numerator d y over denominator d x end fraction equals fraction numerator x plus y over denominator x end fraction$

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parallel

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Statement-II : Every differential equation geometrically represents a family of curve having some common property.

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Statement-II : Every differential equation geometrically represents a family of curve having some common property.

A function f(x) satisfying $stretchy integral subscript 0 end subscript superscript 1 end superscript f left parenthesis t x right parenthesis d t equals n f left parenthesis x right parenthesis comma text end text$ where x>0 , is

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$text If end text y equals fraction numerator x over denominator l n invisible function application vertical line c x vertical line end fraction$ (where c is an arbitrary constant) is the general solution of the differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y over denominator x end fraction plus$ $phi open parentheses fraction numerator x over denominator y end fraction close parentheses text end text$ then the function $text end text phi open parentheses fraction numerator x over denominator y end fraction close parentheses text is : end text$

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