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Correct answer is (b) a1a + b1b + c1c = 0

For explanation I WOULD say: The relation between the plane ax + by +cz + d = 0 and a1, B1, C1 the DIRECTION ratios of a line, if the plane and line are parallel to each other is a1a + b1b + c1c = 0.

Correct option is (c) 66.92

The BEST EXPLANATION: ANGLE between a plane and a LINE sinθ=\(\frac {a1a+b1b+c1c}{\SQRT {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\)

sinθ = 0.92

θ = 66.92

The correct answer is (a) perpendicular

The BEST explanation: θ = 90 degrees

The relation between the PLANE ax + by + cz + d = 0 and a1, B1, c1 the direction ratios of a LINE, if the plane and line are perpendicular to each other is \(\frac {a}{a1} = \frac{b}{b1} = \frac{c}{c1}\).

Correct choice is (a) ORTHOGONAL

Easiest explanation: A plane and A line which are perpendicular to each other or a plane and a line having an ANGLE 90 DEGREES between them are called orthogonal. θ is EQUAL to 90 degrees in sin θ=\(\frac {a1a+b1b+c1c}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\).

Correct choice is (C) \(\frac {a}{a1} = \frac{b}{B1} = \frac{c}{c1}\)

To explain I would SAY: θ = 90 degrees

The relation between the plane ax + by + cz + d = 0 and a1, b1, c1 the DIRECTION RATIOS of a line, if the plane and line are perpendicular to each other is \(\frac {a}{a1} = \frac{b}{b1} = \frac{c}{c1}\).

The correct ANSWER is (d) -7

For explanation: The condition for a plane and a line are PARALLEL to each other is a1a + B1B + C1C = 0.

8(1) + 3(2) + 2(k) = 0

2(k) = -14

k = -7

Correct CHOICE is (b) 3.43

The best I can explain: Angle between a PLANE and a line SIN θ=\(\FRAC {a1a+b1b+c1c}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\)

sinθ = 0.06

θ = 3.43

Correct option is (B) -14

To EXPLAIN: The condition for a plane and a LINE are parallel to each other is a1a + B1B + c1c = 0.

5(3) + 1(-1) + k(1) = 0

K(1) = -14

K = -14

Right choice is (b) PARALLEL

To ELABORATE: The relation between the plane ax + by +cz + d = 0 and a1, B1, c1 the direction ratios of a line, if the plane and line are parallel to each other is a1a + b1b + C1C = 0.

The CORRECT option is (d) Sine

Explanation: The trigonometric function is used to find the angle between a LINE and a plane is sine. If θ is the angle between line whose ratios are A1, b1, c1 and the plane ax + by + CZ + d = 0 then sin θ=\(\frac {a1a2.b1b2.c1c2}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\).

The CORRECT answer is (d) \(cos^{-1}⁡\frac{1}{3\SQRT{3}}\)

Best explanation: The angle between TWO planes of the form \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0 is given by

cos⁡θ=\(\LEFT |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

According to the given question, \(A_1=2,B_1=2,C_1=1 \,and \,A_2=1,B_2=-1,C_2=1\)

cos⁡θ=\(\left |\frac{2(1)+2(-1)+1(1)}{|\sqrt{2^2+2^2+1^2} \sqrt{1^2+(-1)^2+1^2}|}\right |\)

cos⁡θ=\(\left |\frac{1}{\sqrt{9}.\sqrt{3}}\right |\)

θ=\(cos^{-1}⁡\frac{1}{3\sqrt{3}}\).

Right CHOICE is (a) – 29.34

Explanation: Angle between a plane and a line SIN θ=\(\FRAC {a1a+b1b+c1c}{\SQRT {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\)

sin θ = – 0.49

θ = sin^-1(- 0.49)

θ = – 29.34

Right ANSWER is (a) \(cos^{-1}⁡\frac{12}{\sqrt{1302}}\)

To explain: Given that, the NORMAL to the planes are \(\vec{n_1}=4\hat{i}+\hat{j}-2\hat{k} \,and \,\vec{n_2}=5\hat{i}-6\hat{j}+\hat{k}\)

The angle between TWO planes of the form \(\vec{r}.\vec{n_1}=d_1 \,and \,\vec{r}.\vec{n_2}=d_2\) is given by

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

\(|\vec{n_1}|=\sqrt{4^2+1^2+(-2)^2}=\sqrt{21}\)

\(|\vec{n_2}|=\sqrt{5^2+(-6)^2+1^2}=\sqrt{62}\)

\(\vec{n_1}.\vec{n_2}\)=4(5)+1(-6)-2(1)=20-6-2=12

cos⁡θ=\(\frac{12}{\sqrt{21}.\sqrt{62}}=\frac{12}{\sqrt{1302}}\)

∴θ=\(cos^{-1}⁡\frac{12}{\sqrt{1302}}\).

Right option is (b) cos⁡θ=\(\LEFT |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

For EXPLANATION I WOULD say: If the planes are in the Cartesian FORM i.e. \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0, where \(A_1,B_1,C_1 \,and \,A_2,B_2,C_2\) are the direction ratios of the planes, then the angle between them is given by

cos⁡θ=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

Correct CHOICE is (d) (λ+2μ)x+(2λ-μ)y+(3λ-2μ)z=12

Best explanation: GIVEN that the equation of the plane is \(\VEC{r}.[(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}]\)=12

We know that, \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\)

∴\((x\hat{i}+y\hat{j}+z\hat{k}).([(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}])\)=12

⇒(λ+2μ)x+(2λ-μ)y+(3λ-2μ)z=12 is the CARTESIAN equation of the plane.

Correct OPTION is (d) \((\vec{R}-(2\HAT{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0

To elaborate: The position VECTOR of the point (2,1,-1) is \(\vec{a}=2\hat{i}+\hat{j}-\hat{k}\) and the normal vector \(\vec{N}\)perpendicular to the plane is \(\vec{N}=2\hat{i}+\hat{j}-3\hat{k}\)

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

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

Correct OPTION is (b) 2x+y-z=4

The best explanation: Given that the equation of the plane is \(\vec{R}.(2\HAT{i}+\hat{j}-\hat{K})\)=4

We KNOW that, \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\)

∴\((x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+\hat{j}-\hat{k})\)=4

⇒2x+y-z=4 is the Cartesian equation of the plane.

Right ANSWER is (C) p=-1

Explanation: The ANGLE between two lines is given by the EQUATION

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

cos⁡90°=\(\left |\frac{3(1)+p(1)+1(-2)}{\sqrt{3^2+p^2+1^2}.\sqrt{1^2+1^2+(-2)^2}}\right |\)

0=\(|\frac{p+1}{\sqrt{10+p^2}.√6}|\)

0=p+1

p=-1

The CORRECT OPTION is (d) al+mb+cn=d^2

For EXPLANATION I would say: \(\vec{r}.\hat{n}=d\), ax+by+cz=d, lx+my+nz=d are the various ways of REPRESENTING a PLANE.

\(\vec{r}.\hat{n}\)=d is the vector equation of the plane, where \(\hat{n}\) is the unit vector normal to the plane.

ax+by+cz=d, lx+my+nz=d are the Cartesian equation of the plane in the normal form where, a, b, c are the direction ratios and l, m, n are the direction cosines of the normal to the plane respectively.

Right ANSWER is (b) \(\BEGIN{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\)=0

The explanation: If the THREE points are collinear, then they will be in a straight LINE and hence the determinant of the three points will be zero.

i.e.\(\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\)=0

Right answer is (a) perpendicular

For explanation I would SAY: RELATION between the planes A1X + b1y + c1z + d1 = 0 and a2x + B2Y + c21z + d2 = 0, if their normal are parallel to each other is a1 : b1 : c1 = a2 : b2 : C2 ⇒ \(\frac {a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}\).

The correct option is (a) \(\frac{115}{\sqrt{134}}\)

Best explanation: The shortest distance between two lines in CARTESIAN form is given by:

l1:\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\)

l2:\(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)

∴d=\(\left |\frac{\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}}{\sqrt{(b_1 c_2-b_2 c_1)^2+(c_1 a_2-c_2 a_1)^2+(a_1 b_2-a_2 b_1)^2}}\RIGHT |\)

d=\(\left |\frac{\begin{vmatrix}-9&5&2\\2&5&4\\3&6&7\end{vmatrix}}{\sqrt{√(35-24)^2+(12-14)^2+(12-15)^2}}\right |\)

d=\(\left |\frac{-9(35-24)-5(14-12)+2(12-15)}{\sqrt{11^2+2^2+3^2}}\right |\)

d=\(\left |\frac{-99-10-6}{\sqrt{134}}\right |\)

d=\(\frac{115}{\sqrt{134}}\).

Correct answer is (B) 0

The EXPLANATION is: If the direction cosines of a line are l,m,n RESPECTIVELY, then

l^2+m^2+n^2=1

∴\(\FRAC{1}{2}^2+(\frac{\sqrt{3}}{2})^2+x^2=1\)

x^2=\(1-\frac{1}{4}-\frac{3}{4}\)

x^2=0

⇒x=0

Right choice is (b) \(\frac{x+1}{7}=\frac{y+8}{-3}=\frac{z-5}{-3}\)

Best EXPLANATION: The position VECTOR of the GIVEN point is \(\vec{a}=-\hat{i}-8\hat{j}+5\hat{k}\)

The vector which is parallel to the given line is \(7\hat{i}-3\hat{j}-3\hat{k}\)

We know that, \(\vec{R}=\vec{a}+λ\vec{b}\)

∴\(x\hat{i}+y\hat{j}+z\hat{k}=-\hat{i}-8\hat{j}+5\hat{k}+λ(7\hat{i}-3\hat{j}-3\hat{k})\)

=\((-1+7λ) \hat{i}+(-8-3λ) \hat{j}+(5-3λ) \hat{j}\)

⇒\(\frac{x+1}{7}=\frac{y+8}{-3}=\frac{z-5}{-3}\).

Correct CHOICE is (b) 2

The best I can EXPLAIN: 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.

5(1) + 1(1) + 3(-K) = 0

-3k = -6

K = 2

Right choice is (b) \((\vec{r}-(\hat{i}+2\hat{J}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0

Explanation: LET \(\vec{a}=\hat{i}+2\hat{j}-\hat{k}, \,\vec{b}=-\hat{j}+2\hat{k}, \,\vec{c}=3\hat{i}+\hat{j}+\hat{k}\)

The vector equation of the PLANE passing through three points is given by

\((\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]\)=0

\((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[((-\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-\hat{k}))×((3\hat{i}+\hat{j}+\hat{k})–(\hat{i}+2\hat{j}-\hat{k}))]\)=0

\((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0

Right answer is (b) d=\(\sqrt{\FRAC{405}{21}}\)

EXPLANATION: The shortest distance between TWO parallel lines is given by:

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

∴d=\(\left|\frac{(\hat{i}+2\hat{j}-4\hat{K})×(6\hat{i}+2\hat{j}-\hat{k})-(\hat{i}+\hat{j}+\hat{k})}{\sqrt{1^2+2^2+(-4)^2}}\right |\)

=\(\left |\frac{(\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})}{\sqrt{21}} \right |\)

\((\hat{i}+2\hat{j}-4\hat{k})×(5\hat{i}+\hat{j}-2\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}& \hat{k}\\1&2&-4\\5&1&-2\end{vmatrix}\)

=\(\hat{i}(-4+4)-\hat{j}(-2+20)+\hat{k}(1-10)\)

=-\(18\hat{j}-9\hat{k}\)

⇒d=\(\left|\frac{\sqrt{(-18)^2+(-9)^2}}{√21}\right|\)

d=\(\sqrt{\frac{405}{21}}\)

The correct answer is (c) \((2+3λ) \hat{i}+(4λ-3) \hat{j}+(5-2λ)\hat{K}\)

To EXPLAIN: GIVEN that the LINE is passing through the point (2,-3,5). Therefore, the position vector of the line is \(\vec{a}=2\hat{i}-3\hat{j}+5\hat{k}\).

Also given that, the line is parallel to a vector \(\vec{b}=3\hat{i}+4\hat{j}-2\hat{k}\).

We know that, the equation of line passing through a point and parallel to vector is given by \(\vec{r}=\vec{a}+λ\vec{b}\), where λ is a constant.

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

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

Right answer is (C) \(\frac{x-8}{-1}=\frac{y+5}{6}=\frac{z-7}{-3}\)

Easy explanation: CONSIDER A(8,-5,7) and B(7,1,4)

i.e. \((x_1,y_1,z_1)\)=(8,-5,7) and \((x_2,y_2,z_3)\)=(7,1,4)

The cartesian equation for a line passing through two points is given by

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

∴ the cartesian equation for the given line is \(\frac{x-8}{-1}=\frac{y+5}{6}=\frac{z-7}{-3}\)

Right ANSWER is (d) –\(\frac{9}{\SQRT{166}},\frac{-9}{\sqrt{166}},\frac{2}{\sqrt{166}}\)

The best 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(-6,7,3) and Q(3,-2,5)

∴\(PQ=\sqrt{(3+6)^2+(-2-7)^2+(5-3)^2}\)

=\(\sqrt{81+81+4}=\sqrt{166}\)

HENCE, the direction RATIOS are \(l=\frac{-6-3}{\sqrt{166}}=-\frac{9}{\sqrt{166}}\)

m=\(\frac{-2-7}{\sqrt{166}}=\frac{-9}{\sqrt{166}}\)

n=\(\frac{5-3}{\sqrt{166}}=\frac{2}{\sqrt{166}}\)

Right CHOICE is (b) The angle between the planes

Explanation: The angle between the NORMALS to TWO planes is called the angle between the planes. A trigonometric IDENTITY, cosine is used to find the angle called ‘θ’ between two planes.

Right option is (b) 2x+3y+4z=2 and 4x+6y+8z=9

Best EXPLANATION: If TWO planes are of the form \(A_1 x+B_1 y+C_1 z+D_1\)=0 and \(A_2 x+B_2 y+C_2 z+D_2\)=0, then they will be parallel if \(\FRAC{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\)

Consider the planes 2x+3y+4z=2 and 4x+6y+8z=9

\(\frac{A_1}{A_2}=\frac{2}{4}\)

\(\frac{B_1}{B_2}=\frac{3}{6}\)

\(\frac{C_1}{C_2}=\frac{4}{8}\)

\(\frac{2}{4}=\frac{3}{6}=\frac{4}{8}=\frac{1}{2}\)

Hence, the above set of planes are parallel.

The correct answer is (b) \(cos^{-1}⁡⁡\FRAC{43}{\sqrt{3382}}\)

EASY EXPLANATION: The direction ratios are 5, 3, 2 for L1 and 3, 4, 8 for L2

∴ the angle between the TWO lines is given by

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

=\(\frac{15+12+16}{\sqrt{5^2+3^2+2^2}.\sqrt{3^2+4^2+8^2}}\)

=\(\frac{43}{\sqrt{38}.\sqrt{89}}=\frac{43}{\sqrt{3382}}\)

θ=\(cos^{-1}⁡\frac{43}{\sqrt{3382}}\).

The correct choice is (a) \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\)

Explanation: If the LINE is passing through the POINTS (x1, y1, Z1) and has direction COSINES l,m,n of the line, then the cartesian equation of the line is given by

\(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\).

The CORRECT option is (b) \(\frac{5}{\sqrt{90}},\frac{4}{\sqrt{90}},-\frac{7}{\sqrt{90}}\)

For explanation: If a,b,c are the direction ratios and l,m,N are the direction cosines RESPECTIVELY for a given line, then the direction cosines in terms of the direction ratios can be expressed as

l=±\(\frac{a}{\sqrt{a^2+b^2+c^2}}\)

m=±\(\frac{b}{\sqrt{a^2+b^2+c^2}}\)

n=±\(\frac{c}{\sqrt{a^2+b^2+c^2}}\)

Given that, a=5, b=4, c=-7

l=\(\frac{5}{\sqrt{(5^2+4^2+(-7)^2)}}=\frac{5}{\sqrt{(25+16+49)}}=\frac{5}{\sqrt{90}}\)

m=\(\frac{4}{\sqrt{(5^2+4^2+(-7)^2)}}=\frac{4}{\sqrt{90}}\)

n=-\(\frac{7}{\sqrt{(5^2+4^2+(-7)^2)}}=-\frac{7}{\sqrt{90}}\)

The CORRECT answer is (a) The planes are perpendicular to each other

Easy EXPLANATION: We know that if the scalar or dot product of vectors is 0, then they are at right angles to each other. Here the dot product of the NORMAL vectors of the plane is 0, i.e. \(\VEC{n_1}.\vec{n_2}\)=0. Hence, the planes will be perpendicular to each other.

Correct option is (a) \((\vec{r}-(2\HAT{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]\)=0

For explanation I would say: Let \(\vec{a}=2\hat{i}+2\hat{j}, \,\vec{b}=\hat{i}+2\hat{j}+\hat{k}, \,\vec{c}=-\hat{i}+2\hat{j}-2\hat{k}\)

The VECTOR equation of the plane passing through three points is GIVEN by

\((\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]\)=0

\((\vec{r}-(2\hat{i}+2\hat{j})).[((\hat{i}+2\hat{j}+\hat{k})-(2\hat{i}+2\hat{j}))×((-\hat{i}+2\hat{j}-2\hat{k})–(2\hat{i}+2\hat{j}))]\)=0

\((\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]\)=0.

The correct option is (a) \((1+2λ) \hat{i}-(4+5λ) \hat{j}+(4+2λ) \hat{j}\)

To explain: We know that, the EQUATION of a vector passing through a POINT and parallel to another vector is GIVEN by \(\vec{R}=\vec{a}+λ\vec{b}\), where λ is a constant.

The POSITION of vector of the point (1,-4,4) is given by \(\vec{a}=\hat{i}-4\hat{j}+4\hat{k}\)

And \(\vec{b}=2\hat{i}-5\hat{j}+2\hat{k}\)

∴\(\vec{r}=\vec{a}+λ\vec{b}\)

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

=\((1+2λ) \hat{i}-(4+5λ) \hat{j}+(4+2λ) \hat{j}\)

Correct answer is (c) \(\frac{1}{2},-\frac{\SQRT{3}}{2},\frac{1}{\sqrt{2}}\)

Easiest 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=cos⁡60°=\(\frac{1}{2}\)

m=cos⁡β=cos⁡150°=-\(\frac{\sqrt{3}}{2}\)

n=cos⁡γ=cos⁡45°=\(\frac{1}{\sqrt{2}}\).

Right choice is (d) l=\(\FRAC{2}{\sqrt{62}},m=-\frac{3}{\sqrt{62}},N=\frac{7}{\sqrt{62}}\)

Easy explanation: For a given LINE, if a, b, c are the direction ratios and l, m, n are the direction COSINES of the line then

l=±\(\frac{a}{\sqrt{a^2+b^2+c^2}}\)

m=±\(\frac{b}{\sqrt{a^2+b^2+c^2}}\)

n=±\(\frac{c}{\sqrt{a^2+b^2+c^2}}\)

∴l=\(\frac{2}{\sqrt{2^2+(-3)^2+7^2}}, \,m=-\frac{3}{\sqrt{2^2+(-3)^2+7^2}}, \,n=\frac{7}{\sqrt{2^2+(-3)^2+7^2}}\)

HENCE, l=\(\frac{2}{\sqrt{62}}, \,m=-\frac{3}{\sqrt{62}}, \,n=\frac{7}{\sqrt{62}}\).

The correct choice is (d) \(A_1 A_2+B_1 B_2+C_1 C_2\)=0

The BEST I can EXPLAIN: We know that the angle between two planes is given by

cos⁡θ=\(\left |\FRAC{A_1 A_2+B_1 B_2+C_1 C_2}{\SQRT{A_1^2+B_1^2+C_1^2} \sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

Given that, θ=90°

∴cos⁡90°=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

0=\(\left |\frac{A_1 A_2+B_1 B_2+C_1 C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\right |\)

⇒\(A_1 A_2+B_1 B_2+C_1 C_2\)=0.

The CORRECT option is (a) cos θ=\(\frac {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c^2 }}\)

The explanation is: The formula to find the angle between the planes a1x + b1y + C1Z + D1 = 0 and a2x + b2y + c2z + D2 = 0 is cos θ=\(\frac {a1a2+b1b2+c1c2}{\sqrt {a1^2+b1^2+c1^2} \sqrt {a2^2+b2^2+c2^2 }}\). θ is the angle between the normal of two planes.

The correct OPTION is (d) l^2-m^2=n^2-1

Explanation: GIVEN that, a, b, c are the direction RATIOS of the line and l, m, n are the direction cosines of the line,

\(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k\) and l^2+m^2+n^2=1

⇒l=ak, m=BK, n=ck

(ak)^2+(bk)^2+(ck)^2=1

k^2 (a^2+b^2+c^2)=1

k^2=\(\frac{1}{a^2+b^2+c^2}\)

∴k=±\(\frac{1}{\sqrt{(a^2+b^2+c^2)}}\)

Hence, l^2-m^2=n^2-1 is incorrect.