3D Geometry — Page 5 | JEE Mathematics

Q41

If the line,

{{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}

meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from the origin is :

A

{{\sqrt 5 } \over 2}

B 2

\sqrt 5

C 9/2

D 7/2

Correct Answer

Option C

Solution

Let

{{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}

= $\lambda$ Any arbitary point on the line is P( 2 $\lambda$ + 1, 3 $\lambda$ - 1, 4 $\lambda$ + 2).

This point also lies on the plane x + 2y + 3z = 15.

$\therefore$ (2 $\lambda$ + 1) + 2(3 $\lambda$ - 1) + 3(4 $\lambda$ + 2) = 15 $\Rightarrow$ 20 $\lambda$ = 10 $\Rightarrow$ $\lambda$ =

{1 \over 2}

So point P is (2,

{1 \over 2}

, 4). Distance from the origin(O) of point P(2,

{1 \over 2}

, 4) is OP =

\sqrt {{{\left( 2 \right)}^2} + {{\left( {{1 \over 2}} \right)}^2} + {{\left( 4 \right)}^2}}

=

\sqrt {4 + {1 \over 4} + 16}

=

\sqrt {{{81} \over 4}}

=

{9 \over 2}

Q42

The vertices B and C of a

\Delta

ABC lie on the line,

{{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}

such that BC = 5 units. Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :

A 6

B

5\sqrt {17}

C

\sqrt {34}

D

2\sqrt {34}

Correct Answer

Option C

Solution

Area of

\Delta

ABC =

{1 \over 2} \times BC \times AD

Given BC = 5 so we need perpendicular distance of A from line BC.

Let a point D on BC = (3 $\lambda$ - 2, 1, 4 $\lambda$ ) Direction ratio of AD 3 $\lambda$ - 3, 2, 4 $\lambda$ - 2 As AD perpendicular to the BC, so DR of AD.

DR of BC = 0

(3\lambda - 3)3 + 2(0) + (4\lambda - 2)4 = 0

9\lambda - 9 + 16\lambda - 8 = 0 \Rightarrow \lambda = {{17} \over {25}}

Hence,

D = \left( {{1 \over {25}},1,{{68} \over {25}}} \right)

\left| {\overline {AD} } \right| = \sqrt {{{\left( {{1 \over {25}} - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( {{{68} \over {25}} - 2} \right)}^2}}

\Rightarrow \sqrt {{{\left( {{{ - 24} \over {25}}} \right)}^2} + 4 + {{\left( {{{18} \over {25}}} \right)}^2}}

\Rightarrow \sqrt {{{{{\left( {24} \right)}^2} + 4{{\left( {25} \right)}^2} + {{\left( {18} \right)}^2}} \over {{{25}^2}}}}

\Rightarrow \sqrt {{{576 + 2500 + 324} \over {{{25}^2}}}}

\Rightarrow \sqrt {{{3400} \over {{{25}^2}}}} \Rightarrow {{\sqrt {34} .10} \over {25}} = {{2\sqrt {34} } \over 5}

Area of triangle =

{1 \over 2} \times \left| {\overline {BC} } \right| \times \left| {\overline {AD} } \right|

=

{1 \over 2} \times 5 \times {{2\sqrt {34} } \over 5} = \sqrt {34}

Q43

Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :

A 205

\sqrt5

B 63

\sqrt5

C 11/

\sqrt5

D 17/

\sqrt5

Correct Answer

Option C

Solution

P1 : x + y + z – 6 = 0 P2 : 2x + 3y + z + 5 = 0 Equation of plane which passes through the line of intersection of P1 and P2 is P1 + $\lambda$ P2 = 0 $\Rightarrow$ (x + y + z – 6) + $\lambda$ (2x + 3y + z + 5) = 0 $\Rightarrow$ (1 + 2 $\lambda$ )x + (1 + 3 $\lambda$ )y + (1 + $\lambda$ ) + (5 $\lambda$ - 6) = 0 As the above plane is perpendicular to xy plane $\therefore$

\overrightarrow n .\widehat k = 0

$\Rightarrow$

\left[ {\left( {2 + \lambda } \right)\widehat i + \left( {3 + \lambda } \right)\widehat j + \left( {1 + \lambda } \right)\widehat k} \right].\widehat k

= 0 $\Rightarrow$ 1 + $\lambda$ = 0 $\Rightarrow$ $\lambda$ = -1 So, equation of plane - x - 2y - 11 = 0 $\Rightarrow$ x + 2y + 11 = 0 Distance of the point (0, 0, 256) from this plane =

\left| {{{0 + 0 + 11} \over {\sqrt 5 }}} \right|

=

{{11} \over {\sqrt 5 }}

Q44

If the length of the perpendicular from the point (

\beta

, 0,

\beta

) (

\beta

\ne

0) to the line,

{x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}

is

\sqrt {{3 \over 2}}

, then

\beta

is equal to :

A 2

B 1

C -2

D -1

Correct Answer

Option D

Solution

{x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} = p\,\,P\left( {\beta ,0,\beta } \right)

any point on line A = (p, 1, – p – 1) Now, DR of AP $\equiv$ < p – $\beta$ , 1 – 0, – p – 1 – $\beta$ > Which is perpendicular to line so (p – $\beta$ ).

1 + 0.1 – 1(– p – 1 – $\beta$ ) = 0 $\Rightarrow$ p – $\beta$ + p + 1 + $\beta$ = 0

p = {{ - 1} \over 2}

Point

A\left( {{{ - 1} \over 2},1 - {1 \over 2}} \right)

Now, distance AP =

\sqrt {{3 \over 2}}

\Rightarrow A{P^2} = {3 \over 2}

\Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} + 1 + {\left( {\beta + {1 \over 2}} \right)^2} = {3 \over 2}

\Rightarrow 2{\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 2}

\Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 4}

\Rightarrow \beta = 0, - 1,\left( {\beta \ne 0} \right)

\therefore \beta = - 1

Q45

If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the area (in sq. units) of

\Delta

PQR is :

A

{{\sqrt {65} } \over 2}

B

2\sqrt {13}

C

{{\sqrt {91} } \over 2}

D

{{\sqrt {91} } \over 4}

Correct Answer

Option C

Solution

Image of Q in plane

{{\left( {x - 0} \right)} \over 3} = {{\left( {y + 1} \right)} \over { - 1}} = {{z + 3} \over { + 4}} = {{ - 2(1 - 12 - 2)} \over {9 + 1 + 16}} = 1

x = 3, y = –2, z = 1 P(3, –2, 1), Q(0, –1, –3), R(3, –1, –2) Now area of

\Delta

PQR is

{1 \over 2}\left| {\overline {PQ} \times \overline {QR} } \right| = {1 \over 2}\left| {\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 3 & { - 1} & 4 \\ 3 & 0 & 1 \end{array} \right|} \right|

\Rightarrow {1 \over 2}\left| {\left\{ {\widehat i( - 1) - \widehat j(3 - 12) + \widehat k(3)} \right\}} \right|

\Rightarrow {1 \over 2}\sqrt {(1 + 81 + 9)}

\Rightarrow {{\sqrt {91} } \over 2}

Q46

If the plane 2x – y + 2z + 3 = 0 has the distances

{1 \over 3}

and

{2 \over 3}

units from the planes 4x – 2y + 4z +

\lambda

= 0 and 2x – y + 2z +

\mu

= 0, respectively, then the maximum value of

\lambda

+

\mu

is equal to :

A 13

B 9

C 5

D 15

Correct Answer

Option A

Solution

Distance formula (i)

{{\left| {\lambda - 6} \right|} \over {\sqrt {16 + 4 + 16} }} = \left| {{{\lambda - 6} \over 6}} \right| = {1 \over 3}

$\Rightarrow$

\left| {\lambda - 6} \right| = 2

$\Rightarrow$

\lambda = 8,4

(ii)

{{\left| {\mu - 3} \right|} \over {\sqrt {4 + 4 + 1} }} = {2 \over 3}

$\Rightarrow$

\left| {\mu - 3} \right| = 2

$\Rightarrow$

\mu = 5,1

$\therefore$

{\left( {\mu + \lambda } \right)_{\max }} = 13

Q47

The perpendicular distance from the origin to the plane containing the two lines,

{{x + 2} \over 3} = {{y - 2} \over 5} = {{z + 5} \over 7}

and

{{x - 1} \over 1} = {{y - 4} \over 4} = {{z + 4} \over 7},

is :

A

6\sqrt {11}

B

{{11} \over {\sqrt 6 }}

C 11

D 11

\sqrt 6

Correct Answer

Option B

Solution

\left| \begin{array}{lll}i & j & k \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{array} \right|

=

\widehat i

(35 $-$ 28) $-$

\widehat j

(21.7) +

\widehat k

(12 $-$ 5) = 7

\widehat i

$-$ 14

\widehat j

+ 7

\widehat k

=

\widehat i

$-$ 2

\widehat j

+

\widehat k

1(x + 2) $-$ 2(y $-$ 2) + 1 (z + 15) = 0 x $-$ 2y + z + 11 = 0

{{11} \over {\sqrt {4 + 1 + 1} }} = {{11} \over {\sqrt 6 }}

Q48

A perpendicular is drawn from a point on the line

{{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}

to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :

A (4, 0, – 1)

B (2, 0, 1)

C (1, 0, 2)

D (– 1, 0, 4)

Correct Answer

Option B

Solution

{{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1} = \lambda

Let a point P on the line is (2 $\lambda$ + 1, – $\lambda$ –1, + $\lambda$ ) Foot of

{ \bot ^r}Q

is given by

{{x - 2\lambda - 1} \over 1} = {{y + \lambda + 1} \over 1} = {{z - \lambda } \over 1} = - {{\left( {2\lambda - 3} \right)} \over 3}

$\therefore$ Q lies on x + y + z = 3 & x – y + z = 3 $\Rightarrow$ x + z = 3 & y = 0 $\therefore$

y = 0 \Rightarrow \lambda + 1 = {{ - 2\lambda + 3} \over 3} \Rightarrow \lambda = 0

$\therefore$ Q is (2, 0, 1)

Q49

If the line

{{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}}

intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to :

A

2\sqrt 7

B 14

C

2\sqrt {14}

D

\sqrt {14}

Correct Answer

Option C

Solution

{{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}} = \lambda

A(3\lambda + 2,2\lambda - 1, - \lambda + 1)

line on 2x + 3y -z + 13 = 0

\Rightarrow 2(3\lambda + 2) + 3(2\lambda - 1) - ( - \lambda + 1) + 13 = 0

\Rightarrow 13\lambda + 13 = 0 \Rightarrow \lambda = - 1

Now point P(-1, -3, 2) lie on 3x + y + 4z = 16

\Rightarrow 3(3\lambda + 2) + (3\lambda + 2) + 4(3\lambda + 2) = 16

\Rightarrow 9\lambda + 6 + 2\lambda - 4\lambda - 1 + 4 = 16

\Rightarrow 7\lambda = 7 \Rightarrow \lambda = 1

$\Rightarrow$ Q(5, 1, 0) $\therefore$ PQ =

\sqrt {36 + 16 + 4} = \sqrt {56} = 2\sqrt {14}

Q50

A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0, passes through the point :

A (1, –4, 1)

B (1, 4, –1)

C (2, 4, 1)

D (2, –4, 1)

Correct Answer

Option D

Solution

Planes bisecting the given planes are

{{2x - y + 2z - 4} \over 3} = \pm {{x + 2y + 2z - 2} \over 3}

$\Rightarrow$ x - 3y = 2 or 3x + y + 4z = 6 Out of the four given points in question's options only (2, -4, 1) lies on the plane 3x + y + 4z = 6