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401.

If the line with direction ratios k, -4, -1 is perpendicular to the line with direction ratios k, 5, -4 then k =A. `pm 16`B. `pm 2`C. `pm 4`D. `pm 1`

Answer» Correct Answer - C
Discussion
402.

The Cartesian equations ofa line are `6x-2=3y+1=2z-2.`Find its direction ratiosand also find a vector equation of the line.

Answer» The given line is `6x-2=3y+1=2z-2`
To put it in the symmetrical form, we must make the coeffcients `x, y and z` as 1. To do this, we divide each of the expressions in (i) by 6 and obtain `(x-(1//3))/(1)=(y+(1//3))/(2)=(z-1)/(3)`.
This shows that the given line passes through `(1//3, -1//3, 1)` and is parallel to the line whose direction ratios are 1, 2 and 3. Therefore, its vector equation is
`" "vecr=(1)/(3)hati-(1)/(3)hatj+hatk+lamda(hati+2hatj+3hatk)`
Discussion
403.

The direction ratios of AB are -2,2,1. If `A -= (4,1,5)` and l(AB) = 6 units, find coordinates of B.A. (8, 5, 7) or (0, -3, 3)B. (0, 5, 7) or (8, -3, 3)C. (0, 5, 7) or (8, 3, -3)D. (8, 5, 7) or (0, 3, -3)

Answer» Correct Answer - B
Discussion
404.

`A(3,2,0),B(5,3,2),(-9,6,-3)` are the vertices of `/_ ABC` and AD is the bisector of `/_BAC` which meets at D. Find the coordinates of D,

Answer» Correct Answer - `((19)/(8),(57)/(16),(17)/(16))`
Discussion
405.

If the line with direction ratios 5, 3k, 7 is perpendicular to the line with direction ratios k, 1, -8, then k =A. 7B. 8C. 14D. 16

Answer» Correct Answer - A
Discussion
406.

If the coordinates of the points A, B, C, D be `(1, 2, 3)`, `(4, 5, 7)`, `( 4, 3, 6)`and `(2, 9, 2)`respectively, then find the angle between the lines AB and CD.

Answer» Here, `vec(AB) = vecB - vecA = 3hati+3hatj+3hatk`
`vec(CD) = vecD-vecC = 6hati+6hatj+8hatk`
If we see the ratio of the coefficients of `hati, hatj and hatk` for both equations,
`=>3/6 = 3/6 = 4/8 = 1/2`
So, these two vectors are parallel.
So, the angle between `AB` and `CD` will be `0^@` or `180^@`.
Discussion
407.

A line passes through thepoint with position vector `2 hat i-3 hat j+4 hat k`and is in the direction of `3 hat i+4 hat j-5 hat kdot`Find the equations of theline in vector and Cartesian forms.

Answer» Since the line passes through `2hati-3hatj+4hatk` and has direction of `3hati+4hatj-5hatk`, its vector equation is
`" "vecr=hata+lamdahatbrArrvecr=2hati-3hatj+4hatk+lamda(3hati+4hatj-5hatk)`, where `lamda` is a parameter.`" "`(i)
Cartesian equivalent of (i) is `(x-2)/(3)=(y+3)/(4)=(z-4)/(-5)`
Discussion
408.

Consider the line L 1 : x 1 y 2 z 1 312 +++ ==, L2 : x2y2z3 123A. `(12)/(sqrt(65))`B. `(14)/(sqrt(75))`C. `(13)/(sqrt(75))`D. `(13)/(sqrt(65))`

Answer» Correct Answer - c
The plane is given by
`" "-(x+1)- 7(y+2)+5(z+1)=0`
or `" "x+7y-5z+10=0`
`rArr" "` Distance `= (1+7-5+10)/(sqrt(75))= (13)/(sqrt(75))`
Discussion
409.

Consider the line L 1 : x 1 y 2 z 1 312 +++ ==, L2 : x2y2z3 123

Answer» Correct Answer - d
Shortest distance
`= ((1+2)(-1)+(2-2)(-7)+ (1+3)(5))/(5sqrt3) = (17)/(5sqrt3)`
Discussion
410.

Find the equation of the plane through the points (2,2,1) and (9,3,6)` and perpendicular to the plane `2x+6y+6z=1`

Answer» Correct Answer - `3x+4y-5z=9`
Discussion
411.

Find the vector equation ofa line passing through `3 hat i-5 hat j+7 hat k`and perpendicular totheplane `3x-4y+5z=8.`

Answer» The given plane `3x-4y+5z=8 or (3hati-4hatj+5hatk)*(xhati+yhatj+7hatk)=8.`
This shows that `vecd=3hati-4hatj+5hatk` is normal to the given plane.
Therefore, the required line is parallel to `3hati-4hatj+5hatk`.
Since the required line passes through `3hati-5hatj+7hatk`, its equation is given by
`" "vecr=3hati-5hatj+7hatk+lamda(3hati-4hatj+5hatk),` where `lamda` is a parameter.
Discussion
412.

Equation of the panepassing through the points `(2,2,1)a n d(9,3,6),a n d_|_`to the plane `2x+6y+6z-1=0`isa. `3x+4y+5z=9`b. `3x+4y-5z=9`c. `3x+4y-5z=9`d. noneof theseA. `3x+4y+5z=9`B. `3x+4y-5z=9`C. `3x+4y-5z=9`D. none of these

Answer» Correct Answer - b
Any plane through `(2, 2, 1)` is
`" "a(x-2)+b(y-2)+c(z-1)=0`
It passes through `(9, 3, 6)` if `7a+b+5c=0" "` (ii)
Also (i) is perpendicular to `2x+6y+6z-1=0`, we have
`" "2a+6b+6c=0`
`therefore" "a+3b+ 3c=0" "`(iii)
`therefore " "(a)/(-12)= (b)/(-16)= (c)/(20) or (a)/(3)= (b)/(4)= (c)/(-5)" "` [from (ii) and (iii)]
Therefore, the required plane is `3(x-2)+ 4(y-2)-5(z-1)=0 or 3x+4y-5z-9=0`
Discussion
413.

Value of `lamda` such that the line `(x-1)/(2)=(y-1)/(3)=(z-1)/(lamda)` is `bot` to normal to the plane `vecr.(2veci+3vecj+4veck)=0` isA. `-(13)/(4)`B. `-(17)/(4)`C. 4D. none of these

Answer» Correct Answer - a
Since line is parallel to the plane, vector `2hati+3hatj+lamdahatk` is perpendicular to the normal to the plane
`2hati + 3hatj+ 4hatk`
`rArr" "2xx2+ 3xx3 + 4lamda =0`
or `" "lamda = - (13)/(4)`
Discussion