Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

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

< 2, 4, -13 >
< 6, 2, 3 >
< 6/7, 2/7, 3/7 >
None of these

hint Hint:

We are given the coordinates of two points. We have to find the direction cosines of the line joining the two points. Direction cosines are the values of cosines of the given line. They are the cosines of angles it makes with the three axis.

The correct answer is: < 6/7, 2/7, 3/7 >

We will denote the given point by P and Q.
P = (4,3,-5)
Q = (-2,1,-8)
Direction cosines are denoted by l, m, and n.
To find the direction cosines, we take the difference between the coordinates and the distance between two points. It is the ratio of direction ratio and distance between the points.
$fraction numerator x subscript 2 minus x subscript 1 over denominator P Q end fraction comma space fraction numerator y subscript 2 space minus y subscript 1 over denominator P Q end fraction comma fraction numerator z subscript 2 minus z subscript 1 over denominator P Q end fraction$
We will find the distance between the given points using the distance formula.
$P Q space equals square root of left parenthesis x subscript 1 minus x subscript 2 right parenthesis squared plus left parenthesis y subscript 1 space minus space y subscript 2 right parenthesis squared plus left parenthesis z subscript 1 space minus z subscript 2 right parenthesis squared end root space space space space space space space equals square root of left square bracket 4 minus left parenthesis negative 2 right parenthesis right square bracket squared plus left parenthesis 3 minus 1 right parenthesis to the power of 2 end exponent plus left square bracket left parenthesis negative 5 right parenthesis space minus left parenthesis negative 8 right parenthesis right square bracket squared end root space space space space space space space equals square root of left parenthesis 4 space plus space 2 right parenthesis squared space plus space left parenthesis 3 space minus 1 right parenthesis squared plus space left parenthesis negative 5 plus 8 right parenthesis squared end root space space space space space space space equals square root of 6 squared plus 2 squared plus 3 squared end root space space space space space space space space equals square root of 36 plus 4 plus 9 end root space space space space space space space equals square root of 49 space space space space space space space space equals 7 P Q space equals space 7$
Direction ratios are as follows:
x₁ - x₂= 4 - (-2)
= 4 + 2
= 6
y₁ - y₂ = 3 - 2
= 1
z₁ - z₂ = -5 - (-8)
= -5 + 8
= 3
ubstituting the values,
The direction cosines of the line joining two points are as follows:

$l comma space m comma space n space equals space fraction numerator x subscript 1 minus x subscript 2 over denominator P Q end fraction comma fraction numerator y subscript 1 minus y subscript 2 over denominator P Q end fraction comma fraction numerator z subscript 1 minus z subscript 2 over denominator P Q end fraction space space space space space space space space space space space space equals 6 over 7 comma 2 over 7 comma 3 over 7 S o comma space t h e space r i g h t space o p t i o n space i s space less than 6 over 7 comma 2 over 7 comma 3 over 7 greater than$

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

Related Questions to study

The locus of x² + y² + z² = 0 is

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.

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

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

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

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

parallel

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

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

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

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

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

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

parallel

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

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

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

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

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

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.

parallel

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

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

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 =

parallel

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?

The solubility product of a salt having general formula $M X subscript 2 end subscript$ in water is $4 cross times 10 to the power of negative 12 end exponent$ . The concentration of $M to the power of 2 plus end exponent$ ions in the aqueous solution of the salt is:

The solubility product of a salt having general formula $M X subscript 2 end subscript$ in water is $4 cross times 10 to the power of negative 12 end exponent$ . The concentration of $M to the power of 2 plus end exponent$ ions in the aqueous solution of the salt is:

Chemistry-General

the-dissociation of weak electrolyte (weak acid) is expressed in terms of Ostwald dilution law stronger is the acid, weaker is its conjugate base_ The dissociation constants of an acid ( $K subscript a end subscript$ ) and its conjugate base ( $K subscript b end subscript$ ) are related by the given relation: -
$K subscript w end subscript equals K subscript a end subscript cross times K subscript b end subscript$
At $25 to the power of ring operator end exponent C comma K subscript w end subscript$ (Ionic product of water) $equals 10 to the power of negative 14 end exponent$
Phosphoric acid isa weak acid. It is used in fertilizer, food, detergent and toothpaste. Structure of phosphoric acid is:

Aqueous solution of phosphoric acid with a density oflg mL^-1 containing 0.05% by weight of phosphoric acid is used to impart tart taste to many soft drinks. Phosphate ion is an inteife ring radical in qualitative analysis. It should be removed for analysis beyond third group of qualitative analysis
the basicity of phosphoric acid is:

the-dissociation of weak electrolyte (weak acid) is expressed in terms of Ostwald dilution law stronger is the acid, weaker is its conjugate base_ The dissociation constants of an acid ( $K subscript a end subscript$ ) and its conjugate base ( $K subscript b end subscript$ ) are related by the given relation: -
$K subscript w end subscript equals K subscript a end subscript cross times K subscript b end subscript$
At $25 to the power of ring operator end exponent C comma K subscript w end subscript$ (Ionic product of water) $equals 10 to the power of negative 14 end exponent$
Phosphoric acid isa weak acid. It is used in fertilizer, food, detergent and toothpaste. Structure of phosphoric acid is:

Aqueous solution of phosphoric acid with a density oflg mL^-1 containing 0.05% by weight of phosphoric acid is used to impart tart taste to many soft drinks. Phosphate ion is an inteife ring radical in qualitative analysis. It should be removed for analysis beyond third group of qualitative analysis
the basicity of phosphoric acid is:

Chemistry-General

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.