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Right option is (d) \(\frac{57}{\sqrt{477}}\)

To EXPLAIN I would SAY: We know that, the shortest distance between two skew lines is given by

d=\(\LEFT |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)

The VECTOR EQUATIONS of the two lines is

\(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+λ(3\hat{i}+2\hat{j}+5\hat{k})\)

\(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+μ(3\hat{i}-2\hat{j}+4\hat{k})\)

∴d=\(\left|\frac{((3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k}).(4\hat{i}-\hat{j}+5\hat{k})-(2\hat{i}+2\hat{j}-2\hat{k}))}{|(3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})|}\right |\)

\((3\hat{i}+2\hat{j}+5\hat{k})×(3\hat{i}-2\hat{j}+4\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&2&5\\3&-2&4\end{vmatrix}\)

=\(\hat{i}(8+10)-\hat{j}(12-15)+\hat{k}(-6-6)\)

=\(18\hat{i}+3\hat{j}-12\hat{k}\)

d=\(\left|\frac{(18\hat{i}+3\hat{j}-12\hat{k}).(2\hat{i}-3\hat{j}+7\hat{k})}{\sqrt{18^2+3^2+(-12)^2}}\right |\)

d=\(\left|\frac{36-9-84}{\sqrt{477}}\right |\)=\(\frac{57}{\sqrt{477}}\).

Right choice is (C) \(\sqrt{\frac{171}{14}}\)

Easy explanation: The DISTANCE between two parallel lines is given by

d=\(\left |\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |\)

=\(\left |\frac{((3\hat{i}-2\hat{J}+\hat{k})×((3\hat{i}+2\hat{j}-\hat{k})-(2\hat{i}-\hat{j}+\hat{k})))}{|\sqrt{3^2+(-2)^2+1^2}|}\right |\)

=\(\left |\frac{(3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k}))}{\sqrt{14}}\right |\)

\( (3\hat{i}-2\hat{j}+\hat{k})×(\hat{i}+3\hat{j}-2\hat{k})=\BEGIN{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-2&1\\1&3&-2\end{vmatrix}\)

=\(\hat{i}(4-3)-\hat{j}(-6-1)+\hat{k}(9+2)\)

=\(\hat{i}+7\hat{j}+11\hat{k}\)

∴d=\(\frac{|\hat{i}+7\hat{j}+11\hat{k}|}{\sqrt{14}}=\frac{\sqrt{1+49+121}}{\sqrt{14}}=\sqrt{\frac{171}{14}}\).

Right ANSWER is (C) l=-\(\frac{1}{2}, \,m=\frac{1}{\sqrt{2}}, \,n=\frac{\sqrt{3}}{2}\)

Easy EXPLANATION: Let l, m, n be the direction cosines of the line.

We know that, if α, β, γ are the angles that the line MAKES with the x, y, z- axis respectively, then

l=cos⁡α

m=cos⁡β

n=cos⁡γ

∴l=cos⁡120°, m=cos⁡45°, n=cos⁡30°

Hence, \(l=-\frac{1}{2}, \,m=\frac{1}{\sqrt{2}}, \,n=\frac{\sqrt{3}}{2}\)

The CORRECT answer is (a) Orthogonal

The best I can EXPLAIN: The PLANES which are PERPENDICULAR to each other i.e.; having an angle 90 degrees between them are called orthogonal planes. HENCE, Orthogonal planes have an angle 90 degrees between them.

Right option is (b) True

To explain: A mathematical symbol θ is used to FIND the ANGLE between line and a NORMAL line to the plane π along with a trigonometric function called sine. Hence, the formula

sin θ=\(\frac {a1a+b1b+c1c}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\).

The correct ANSWER is (C) perpendicular

Easiest explanation: θ = 90 degrees ⇒ cos θ

a1a2 + b1b2 – c1c2 = 0

Relation between the PLANES a1x + b1y + c1z + d1 = 0 and a2x + B2Y + c21z + d2 = 0, if their NORMAL are perpendicular to each other is a1a2 + b1b2 + c1c2 = 0.

The CORRECT option is (a) 4x-2y+5z+7=0

Explanation: The POSITION VECTOR of the point (3,2,-3) is \(\vec{a}=3\hat{i}+2\hat{j}-3\hat{k}\) and the normal vector \(\vec{N}\) perpendicular to the plane is \(\vec{N}=4\hat{i}-2\hat{j}+5\hat{k}\)

Therefore, the vector equation of the plane is given by \((\vec{r}-\vec{a}).\vec{N}\)=0

Hence, \((\vec{r}-(3\hat{i}+2\hat{j}-3\hat{k})).(4\hat{i}-2\hat{j}+5\hat{k})\)=0

4(x-3)-2(y-2)+5(z+3)=0

4x-2y+5z+7=0.

The correct option is (a) cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)

For explanation I would SAY: If two planes of the FORM \(\vec{r}.\vec{n_1}=d_1\) and \(\vec{r}.\vec{n_2}=d_2\) where \(\vec{n_1} \,and \,\vec{n_2}\) are the normals to the plane, then the angle between them is GIVEN by

cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)

The correct answer is (a) d=\(\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)

EASY explanation: The DISTANCE between TWO lines L1 and l2 with the equations

\(\vec{r}=\vec{a_1}+λ\vec{b_1}\)

\(\vec{r}=\vec{a_2}+μ\vec{b_2}\)

Then, the distance between the two lines is GIVEN by the formula

d=\(\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)

The correct option is (c) cos⁡θ=\(\left |\frac{\VEC{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |\)

The best explanation: Given that the EQUATIONS of the LINES are

\(\vec{r}=\vec{a_1}+λ\vec{b_1} \,and \,\vec{r}=\vec{a_2}+μ\vec{b_2}\)

∴ the ANGLE between the TWO lines is given by

cos⁡θ=\(\left |\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |\).

Correct choice is (c) AX + by + CZ+ d = 0

Explanation: The angle between a line and a plane is the complement of the angle between the line L and a NORMAL line to the plane π. If θ is the angle between line WHOSE ratios are a1, b1, c1 and the plane ax + by + cz + d = 0 then SIN θ=\(\frac {a1a+b1b+c1c}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\).

Right option is (c) –\(\frac{5}{\sqrt{270}}, \frac{7}{\sqrt{270}},\frac{14}{\sqrt{270}}\)

For explanation: The direction cosines of two lines passing through two points is given by:

\(\frac{x_2-x_1}{PQ}, \frac{y_2-y_1}{PQ}, \frac{z_2-z_1}{PQ} \)

and \(PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

In the given problem we have, P(4,-5,-6) and Q(-1,2,8)

∴\(PQ = \sqrt{(-1-4)^2+(2+5)^2+(8+6)^2}\)

\(=\sqrt{25+49+196}=\sqrt{270}\)

HENCE, the direction RATIOS are \(l=\frac{(-1-4)}{\sqrt{270}}=-\frac{5}{\sqrt{270}}\)

\(m=\frac{(2+5)}{\sqrt{270}}=\frac{7}{\sqrt{270}}\)

\(n=\frac{(8+6)}{\sqrt{270}}=\frac{14}{\sqrt{270}}\).

The CORRECT answer is (c) \(θ=cos^{-1}⁡\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |\)

For explanation I would say: If L1 and L2 have the direction RATIOS \(a_1,b_1,c_1 \,and \,a_2,b_2,c_2\) respectively then the angle between the LINES is given by

\(cos⁡θ=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |\)

\(θ=cos^{-1}⁡\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |\)

The correct answer is (c) \(cos^{-1}⁡\frac{1}{\sqrt{132}}\)

The EXPLANATION: Given that, the normal to the PLANES are \(\vec{n_1}=2\hat{i}-\hat{J}+\hat{K}\) and \(\vec{n_2}=3\hat{i}+2\hat{j}-3\hat{k}\)

cos⁡θ=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)

\(|\vec{n_1}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}\)

\(|\vec{n_2}|=\sqrt{3^2+2^2+(-3)^2}=\sqrt{22}\)

\(\vec{n_1}.\vec{n_2}\)=2(3)-1(2)+1(-3)=6-2-3=1

cos⁡θ=\(\frac{1}{\sqrt{22}.\sqrt{6}}=\frac{1}{\sqrt{132}}\)

∴θ=\(cos^{-1}⁡\frac{1}{\sqrt{132}}\).

Right ANSWER is (b) a1a2 + b1b2 + C1C2 = 0

Explanation: θ = 90 degrees ⇒ COS θ

a1a2 + b1b2 – c1c2 = 0

Relation between the planes a1x + b1y + c1z + D1 = 0 and a2x + b2y + c21z + D2 = 0, if their normal are perpendicular to each other is a1a2 + b1b2 + c1c2 = 0.

Right choice is (a) θ=\(cos^{-1}\FRAC{⁡20}{\SQRT{602}}\)

The explanation is: If two lines have the equations \(\vec{r}=\vec{a_1}+λ\vec{b_1} \,and \,\vec{r}=\vec{a_2}+μ\vec{b_2}\)

Then, the angle between the two lines will be given by

cos⁡θ=\(\LEFT |\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |\)

=\(\left |\frac{(\hat{i}-2\hat{j}+3\hat{K}).(5\hat{i}-3\hat{j}+3\hat{k})}{\sqrt{1^2+(-2)^2+(3)^2).√(5^2+(-3)^2+3^2}}\right |\)

=\(\frac{5+6+9}{√14.√43}=\frac{20}{√602}\)

θ=\(cos^{-1}⁡\frac{20}{\sqrt{602}}\)

The correct CHOICE is (b) The angle between a LINE and a plane

Easiest explanation: The angle between a line and a plane is the complement of the angle between the line L and a normal line to the plane π. If θ is the angle between line whose ratios are a1, b1, c1 and the plane AX + by + cz + d = 0 then SIN θ=\(\frac {a1a+b1b+c1c}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\).

The correct CHOICE is (B) Cosecant

To ELABORATE: The SYMBOL ‘θ’ represents the angle between two planes. A trigonometric function called cosine is used the find the angle i.e.; θ between the normal of two planes.

The correct answer is (b) \(\vec{r}.(\frac{2\hat{i}}{\SQRT{38}}+\frac{3\hat{J}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)

The best I can EXPLAIN: Let \(\vec{n}=2\hat{i}+3\hat{j}-5\hat{k}\)

\(\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{(2^2+3^2+(-5)^2)}}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{38}}\)

HENCE, the required equation of the PLANE is \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)

Correct choice is (d) 11

Easy explanation: Relation between the the planes a1x + b1y + c1z + D1 = 0 and a2x + b2y + c21z + D2 = 0, if their NORMAL are perpendicular to each other is A1A2 + b1b2 + c1c2 = 0.

1(3) + 2(4) + k(-1) = 0

k(-1) = -11

k = 11

The correct answer is (b) 34.91

To explain: Angle between two PLANES COS θ=\(\FRAC {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c2^2 }}\)

cos θ = 0.82

θ = 34.91

Correct choice is (B) True

Explanation: The formula to find angle between the NORMAL of two PLANES is

cos θ=\(\FRAC {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt{a2^2+b2^2+c2^2 }}\) not cos θ=\(\frac {a1a2.b1b2.c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c2^2 }}\) because the numerator should contain SUM of co-efficients not their product.

Right option is (c) \(\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})\)=4 and \(\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5

The explanation is: Consider the SET of PLANES \(\vec{r}.(2\hat{i}-2\hat{j}+\hat{k})=4 \,and \,\vec{r}.(\hat{i}+2\hat{j}+2\hat{k})\)=5

For the set of planes to be PERPENDICULAR \(\vec{n_1.}\vec{n_2}\)=0

In the above set of planes, \(\vec{n_1}=2\hat{i}-2\hat{j}+\hat{k}\) and \(\vec{n_2}=\hat{i}+2\hat{j}+2\hat{k}\)

∴\(\vec{n_1}.\vec{n_2}\)=2(1)-2(2)+1(2)=0

Hence, they are perpendicular.

The CORRECT choice is (c) p=13

To EXPLAIN: We know that, if TWO lines are perpendicular to each other then,

\(a_1 a_2+b_1 b_2+c_1 c_2=0\)

i.e.4(p)+(-2)2+4(-12)=0

4p-4-48=0

4p=52

p=\(\FRAC{52}{4}\)=13.

Correct ANSWER is (d) cos⁡θ=\(\left |l_1 \,l_2+m_1 \,m_2+n_1 \,n_2\right |\)

For EXPLANATION: If two lines L1 and L2 are having direction cosines \(l_1,m_1,n_1 \,and \,l_2,m_2,n_2\) respectively, then the ANGLE between the lines is given by

cos⁡θ=\(\left |l_1 \,l_2+m_1 \,m_2+n_1 \,n_2\right |\)

The CORRECT choice is (d) \((1+5λ) \hat{i}-3λ\hat{J}+(4-3λ) \hat{k}\)

Best explanation: Consider the points A(1,0,4) and B(6,-3,1)

Let \(\VEC{a} \,and \,\vec{b}\) be the position vectors of the points A and B.

∴\(\vec{a}=\hat{i}+4\hat{k}\)

\(\vec{b}=6\hat{i}-3\hat{j}+\hat{k}\)

∴\(\vec{r}=\hat{i}+4\hat{k}+λ(6\hat{i}-3\hat{j}+\hat{k}-(\hat{i}+4\hat{k}))\)

=\(\hat{i}+4\hat{k}+λ(5\hat{i}-3\hat{j}-3\hat{k})\)

=\((1+5λ) \hat{i}-3λ\hat{j}+(4-3λ) \hat{k}\)

Right answer is (d) – 8

Easiest explanation: If their normals are PERPENDICULAR to each other then A1A2 + b1b2 + C1C2 = 0.

2(3) + 2(1) + s(1) = 0

s(1) = – 8

k = – 8

The CORRECT answer is (b) \(\frac{x-1}{3}=\frac{y+9}{8}=\frac{z-8}{-2}\)

Easiest explanation: The position vector for the point A(1,-9,8) and B(4,-1,6)

\(\vec{a}=\hat{i}-9\hat{j}+8\hat{k}\)

\(\vec{b}=4\hat{i}-\hat{j}+6\hat{k}\)

∴\(\vec{r}=\vec{a}+λ(\vec{b}-\vec{a})\)

The above vector equation can be expressed in CARTESIAN FORM as:

\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

∴ The cartesian equation for the given line is \(\frac{x-1}{3}=\frac{y+9}{8}=\frac{z-8}{-2}\)

Correct choice is (c) \(\FRAC{18}{\sqrt{10}}\)

To elaborate: The distance between two SKEW lines is given by

d=\(\left |\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)

\(\vec{r}=3\hat{i}-4\hat{J}+2\hat{k}+λ(4\hat{i}+\hat{j}+\hat{k})\)

\(\vec{r}=5\hat{i}+\hat{j}-\hat{k}+μ(2\hat{i}-\hat{j}-3\hat{k})\)

d=\(\left |\frac{((4\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}-\hat{k})).((3\hat{i}-4\hat{j}+2\hat{k})-(2\hat{i}-\hat{j}-\hat{k}))}{|4\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}-3\hat{k})|}\right |\)

\((4\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}-\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4&1&1\\2&-1&-1\end{vmatrix}\)

=\(\hat{i}(-1+1)-\hat{j}(-4-2)+\hat{k}(-4-2)\)

=\(6\hat{j}-6\hat{k}\)

d=\(\left |{6\hat{j}-6\hat{k}).(\hat{i}-3\hat{j}+3\hat{k})}{|6\hat{i}-2\hat{k}|}\right |\)

\(\left|\frac{0-18-18}{\sqrt{6^2+2^2}}\right |=\frac{36}{\sqrt{40}}=\frac{18}{\sqrt{10}}\)

Correct choice is (c) \(x_2-x_1,y_2-y_1,z_2-z_1\)

Easiest explanation: Let a.,B,c be the direction RATIOS of the line segment PQ.

Then, the direction ratios of the line segment joining \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\) is given by

\(a=x_2-x_1\)

\(b=y_2-y_1\)

\(c=z_2-z_1\)

Correct CHOICE is (d) 54.54

Explanation: Angle between two planes cos cos θ=\(\FRAC {a1a2+b1b2+c1c2}{\SQRT {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c2^2 }}\)

cos θ = 0.58

θ = cos^-1(0.58)

θ = 54.54

Right answer is (d) \((-5+9λ) \HAT{i}+(3-6λ)\hat{j}+(1+λ) \hat{K}\)

Explanation: Consider the POINTS A(-5,3,1) and B(4,-3,2)

LET \(\vec{a} \,and\, \vec{b}\) be the position vectors of the points A and B.

∴\(\vec{a}=-5\hat{i}+3\hat{j}+\hat{k}\)

\(\vec{b}=4\hat{i}-3\hat{j}+2\hat{k}\)

∴\(\vec{r}=-5\hat{i}+3\hat{j}+\hat{k}+λ(4\hat{i}-3\hat{j}+2\hat{k}-(-5\hat{i}+3\hat{j}+\hat{k}))\)

=-\(5\hat{i}+3\hat{j}+\hat{k}+λ(9\hat{i}-6\hat{j}+\hat{k})\)

=\((-5+9λ) \hat{i}+(3-6λ)\hat{j}+(1+λ) \hat{k}\)