Write the sum of intercepts cut off by the plane on the three a

Find the coordinates of the foot of perpendicular drawn from the point A
(-1,8,4) to the line joining the points B(0,-1,3) and C(2,-3,-1). Hence find the image of the point A in the line BC.


Suppose P be the foot of the perpendicular drawn from point A on the line joining
points B and C.
Let P’ (a, b, c) be the coordinates of the image of point A.
fraction numerator straight x minus straight x subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction space equals space fraction numerator straight y minus straight y subscript 1 over denominator straight y subscript 2 minus end subscript straight y subscript 1 end fraction space space equals space fraction numerator straight z minus straight z subscript 1 over denominator straight z subscript 2 minus straight z subscript 1 end fraction

fraction numerator straight x minus 0 over denominator 2 end fraction space equals space fraction numerator space straight y plus 1 over denominator negative 2 end fraction space equals space fraction numerator straight z minus 3 over denominator negative 4 end fraction space equals space straight lambda

General space coordinates space of space straight P space is space left parenthesis 2 straight lambda comma negative 2 straight lambda minus 1 comma negative 4 straight lambda plus 3 right parenthesis

Direction space ratio space of space AP space left parenthesis 2 straight lambda plus 1 comma negative 2 straight lambda minus 9 comma 4 straight lambda minus 1 right parenthesis

AP perpendicular BC

2 left parenthesis 2 straight lambda plus 1 right parenthesis minus 2 left parenthesis negative 2 straight lambda minus 9 right parenthesis minus 4 left parenthesis negative 4 straight lambda minus 1 right parenthesis equals 0

4 straight lambda plus 2 plus 4 straight lambda plus 18 plus 16 straight lambda plus 4 space equals 0

24 plus 24 straight lambda space equals 0

straight lambda equals negative 1

straight P left parenthesis negative 2 comma 1 comma 7 right parenthesis

Coordinates space of space foot space of space perpendicular space is space left parenthesis negative 2 comma 1 comma 7 right parenthesis

Coordinates space of space image space of space straight A space is space straight P to the power of apostrophe left parenthesis space straight a comma straight b comma straight c right parenthesis space is space

fraction numerator straight a minus 1 over denominator 2 end fraction equals negative 2 comma space straight a equals negative 3

fraction numerator straight b plus 8 over denominator 2 end fraction space equals space 1 comma space straight b equals negative 6

fraction numerator straight c plus 4 over denominator 2 end fraction space equals space 7 comma space straight c equals space 10

straight P apostrophe space left parenthesis negative 3 comma negative 6 comma 10 right parenthesis

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Find the distance of a point (2, 5, −3) from the plane straight r with rightwards arrow on top. open parentheses 6 straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top close parentheses space equals 4.


Consider the vector equation of the plane.
straight r with rightwards arrow on top. space open parentheses 6 straight i with hat on top minus 3 straight j with hat on top space plus space 2 straight k with hat on top close parentheses space equals space 4
space space space space space rightwards double arrow open parentheses straight x straight i with hat on top space plus space straight y straight j with hat on top space plus space straight z straight k with hat on top close parentheses. space open parentheses 6 straight i with hat on top minus 3 straight j with hat on top space plus space 2 straight k with hat on top close parentheses space equals space 4
space space space space space rightwards double arrow 6 straight x minus 37 plus 2 straight z space equals 4
space space space space space rightwards double arrow 6 straight x minus 3 straight y plus 2 straight z minus 4 space equals space 0 space space space space space
space space
Thus the Cartesian equation of the plane is
6x-3y+2z-4 = 0
Let d be the distance between the point (2, 5, -3)
Thus space straight d space equals space open vertical bar fraction numerator ax subscript 1 plus by subscript 1 plus cz subscript 1 plus straight d over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction close vertical bar
space space space rightwards double arrow straight d space equals space open vertical bar fraction numerator 6 cross times 2 minus 3 cross times 5 plus 2 cross times left parenthesis negative 3 right parenthesis minus 4 over denominator square root of 6 squared plus left parenthesis negative 3 right parenthesis squared plus 2 squared end root end fraction close vertical bar
space space space rightwards double arrow straight d space equals space open vertical bar fraction numerator 12 minus 15 minus 6 minus 4 over denominator square root of 36 plus 9 plus 4 end root end fraction close vertical bar
space space space space rightwards double arrow straight d space equals space open vertical bar fraction numerator negative 13 over denominator square root of 49 end fraction close vertical bar
space space space space rightwards double arrow straight d space equals space 13 over 7 units

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Find the equation of the plane which contains the line of intersection of the planes
straight r with rightwards arrow on top. left parenthesis straight i with hat on top minus 2 straight j with hat on top plus 3 straight k with hat on top right parenthesis minus 4 space equals space 0
and
straight r with rightwards arrow on top. left parenthesis negative 2 straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis plus 5 space space equals space 0
and whose intercept on the x-axis is equal to that of on y-axis.


stack straight r. with rightwards arrow on top left parenthesis straight i with hat on top minus 2 straight j with hat on top plus 3 straight k with hat on top right parenthesis minus 4 space equals 0

straight r with rightwards arrow on top. left parenthesis negative 2 straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis plus 5 space equals 0

straight r with rightwards arrow on top space left parenthesis straight i with hat on top minus 2 straight j with hat on top plus 3 straight k with hat on top right parenthesis plus straight lambda open curly brackets straight r with rightwards arrow on top. left parenthesis negative 2 straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis close curly brackets minus 4 plus 5 straight lambda space equals 0

rightwards double arrow space straight r with rightwards arrow on top left square bracket left parenthesis 1 minus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis negative 2 plus straight lambda right parenthesis straight j with hat on top space plus left parenthesis 3 plus straight lambda right parenthesis straight k with hat on top right square bracket minus 4 plus 5 straight lambda equals 0

left parenthesis 1 minus 2 straight lambda right parenthesis space straight x space plus left parenthesis negative 2 plus straight lambda right parenthesis straight y plus left parenthesis 3 plus straight lambda right parenthesis straight z space equals negative 5 straight lambda plus 4

rightwards double arrow space fraction numerator straight x over denominator begin display style fraction numerator negative 5 straight lambda plus 4 over denominator 1 minus 2 straight lambda end fraction end style end fraction plus fraction numerator straight y over denominator begin display style fraction numerator negative 5 straight lambda plus 4 over denominator negative 2 plus straight lambda end fraction end style end fraction plus fraction numerator straight z over denominator begin display style fraction numerator negative 5 straight lambda plus 4 over denominator 3 plus straight lambda end fraction end style end fraction equals 1

therefore space fraction numerator negative 5 straight lambda plus 4 over denominator 1 minus 2 straight lambda end fraction space equals fraction numerator negative 5 straight lambda plus 4 over denominator negative 2 plus straight lambda end fraction

rightwards double arrow space 1 minus 2 straight lambda space equals negative 2 plus straight lambda

rightwards double arrow space minus 3 straight lambda space equals space minus 3

rightwards double arrow space straight lambda space equals 1

therefore space equation space of space the space required space plane
minus straight x minus straight y plus 4 straight z equals 1
straight x plus straight y minus 4 straight z minus 1 space equals space 0

vector space equation space of space the space required space plane

equals straight r with rightwards arrow on top. left parenthesis straight i with hat on top plus straight j with hat on top minus 4 straight k with hat on top right parenthesis minus 1 space equals 0
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Write the sum of intercepts cut off by the planestraight r with rightwards arrow on top. open parentheses 2 straight i with hat on top plus straight j with hat on top minus straight k with hat on top close parentheses minus 5 space equals 0 on the three axes.


Given,
straight r with rightwards arrow on top. open parentheses 2 straight i with hat on top plus straight j with hat on top minus straight k with hat on top close parentheses minus 5 equals 0
in space cartesian space form
2 straight x space plus straight y minus straight z minus 5 space equals 0
2 straight x plus straight y minus straight z equals 5
fraction numerator 2 straight x over denominator 5 end fraction plus straight y over 5 minus straight z over 5 equals 1
fraction numerator straight x over denominator 5 divided by 2 end fraction plus straight y over 5 plus fraction numerator straight z over denominator negative 5 end fraction equals 1
Intercept space cutt space of space on space the space axes space open parentheses 5 over 2 comma 5 comma negative 5 close parentheses
straight x over straight a plus straight y over straight b plus straight z over straight c equals 1
straight a equals 5 over 2 space straight b equals 5 space straight c equals negative 5
straight a plus straight b plus straight c equals 5 divided by 2

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Find x such that the four points A(4, 1, 2), B(5, x, 6) , C(5, 1, -1) and D(7, 4, 0) are coplanar.


Position space vector space of space OA with rightwards arrow on top space equals space 4 straight i with hat on top plus straight j with hat on top space plus space 2 straight k with hat on top
Position space vector space of space OB with rightwards arrow on top space equals space 5 straight i with hat on top plus straight x straight j with hat on top plus 6 straight k with hat on top
Position space vector space of space OD with rightwards arrow on top space equals 7 straight i with hat on top plus 4 straight j with hat on top plus 0 straight k with hat on top
AB with rightwards arrow on top space equals space OB with rightwards arrow on top space minus space OA with rightwards arrow on top
space space equals space 5 straight i with hat on top plus straight x straight j with hat on top plus 6 straight k with hat on top minus 4 straight i with hat on top minus straight j with hat on top minus 2 straight k with hat on top equals straight i with hat on top plus left parenthesis straight x minus 1 right parenthesis straight j with hat on top plus 4 straight k with hat on top
space space space space AC with rightwards arrow on top space equals space OC with rightwards arrow on top space minus space OA with rightwards arrow on top
space space space space space space equals 5 straight i with hat on top space plus straight j with hat on top minus straight k with hat on top space minus 4 straight i with hat on top minus straight j with hat on top minus 2 straight k with hat on top space equals space straight i with hat on top minus 3 straight k with hat on top
space space space space space space AD with rightwards arrow on top space equals space OD with rightwards arrow on top space minus space OA with rightwards arrow on top
space space space space space space space space equals 7 straight i with hat on top plus 4 straight j with hat on top plus 0 straight k with hat on top minus 4 straight i with hat on top minus straight j with hat on top minus 2 straight k with hat on top equals 3 straight i with hat on top plus 3 straight j with hat on top minus 2 straight k with hat on top
The above three vectors are coplanar
rightwards double arrow AB with rightwards arrow on top. space open parentheses AC with rightwards arrow on top space cross times space AD with rightwards arrow on top close parentheses space equals space 0
rightwards double arrow space open vertical bar table row 1 cell straight x minus 1 end cell cell space space 4 end cell row 1 0 cell negative 3 end cell row 3 3 cell negative 2 end cell end table close vertical bar space equals space 0
rightwards double arrow space 1 left parenthesis 10 plus 9 right parenthesis minus left parenthesis straight x minus 1 right parenthesis space left parenthesis negative 2 plus 9 right parenthesis plus 4 left parenthesis 3 minus 0 right parenthesis space equals space 0
rightwards double arrow 9 minus 7 left parenthesis straight x minus 1 right parenthesis plus 12 space equals 0
rightwards double arrow negative 7 left parenthesis straight x minus 1 right parenthesis space equals space minus 21
rightwards double arrow straight x minus 1 space equals space 3
therefore space space straight x space equals space 4
 
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