Mathematics Test - 22

CONCEPT:

The equation of a line with direction ratio  that passes through the point (x1, y1, z1) is given by the formula:

\(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

CALCULATION:

Given: Equation of line is 2x = 3y = 5 - 4z

We will first be converting the above expression in standard form for comparison, i.e. we need to get rid of coefficients of x, y, and z, i.e. 2, 3, and 4 respectively.

LCM of 2, 3, and 4 is 12.

∴ Dividing the above equation by 12, we get:

\(⇒ \frac{{2x}}{{12}} = \frac{{3y}}{{12}} = \frac{{5\ -\ 4z}}{{12}}\;\)

\(⇒ \frac{x}{6} = \frac{y}{4} = \frac{{z - \frac{5}{4}}}{{ - 3}}\;\)

By comparing the above equation with \(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\) we get

⇒ a = 6, b = 4 and c = -3

So, the direction ratios of the given line is: <6, 4, - 3>

Hence, the correct option is 2.

Concept:

The vector form of the component of \(\vec{a}\) along \(\vec{b}\) = \(\rm \left(\dfrac {\vec a. \vec b}{|\vec b|^2}\right)\vec b\)

Calculations:

Given,  \(\vec{a} = 4\hat{j}\) and \(\vec{b} = 3\hat{j} + 4\hat{k}\)

⇒ \(\rm \vec a . \vec b = (4\vec j).(3\vec j + 4 \vec k)\)

⇒ \(\rm \vec a . \vec b = 12\)

and  \(\rm |\vec b| = 5 \)

The vector form of the component of \(\vec{a}\) along \(\vec{b}\) = \(\rm \left(\dfrac {\vec a. \vec b}{|\vec b|^2}\right)\vec b\)

= \(\rm \dfrac {12}{25}(3\vec j + 4 \vec k)\)

Hence, if \(\vec{a} = 4\hat{j}\) and \(\vec{b} = 3\hat{j} + 4\hat{k}\), then the vector form of the component of \(\vec{a}\) along \(\vec{b}\) is \(\rm \dfrac {12}{25}(3\vec j + 4 \vec k)\)

Concept:

1. If two vector a and b then \(\left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \left| {{\rm{\vec a}}} \right|{\rm{\;}}\left| {{\rm{\vec b}}} \right|\sin {\rm{\theta }}\)

 Where θ is angle between vector a and b.

2. Vector A = \({\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\) then |A| = \(\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2}} \)

Calculation:

Given that,

\(\overrightarrow {|{\rm{a|}}} { = 2,{\rm{\;}}\overrightarrow {|{\rm{b}|}} }= 7​​​​\:and\:\overrightarrow {{\rm{a\;}}} \times {\rm{\;\vec b}} = 3{\rm{\hat i}} + 2{\rm{\hat j}} + 6{\rm{\hat k}},\)

Now

\(\begin{array}{l} \Rightarrow \left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \sqrt {{3^2} + {2^2} + {6^2}} \\ \Rightarrow \left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \sqrt {49} = 7\\ \Rightarrow \left| {{\rm{\vec a\;}} \times {\rm{\;\vec b}}} \right| = \left| {{\rm{\vec a}}} \right|{\rm{\;}}\left| {{\rm{\vec b}}} \right|\sin {\rm{\theta }} \end{array}\)

Put all values in above equation

⇒ 7 = 2 × 7 × sin θ

⇒ sin θ = 1/2

CONCEPT:

Let A (x1, y1) and B (x2, y2) be the two given points and the point P (x, y) divide the line joining the points A and B in the ratio m : n, then

• Point of internal division is given as: \(\left( {x,\;y} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
• Point of external division is given as: \(\left( {x,y} \right) = \left( {\frac{{m{x_2} - n{x_1}}}{{m - n}},\frac{{m{y_2} - n{y_1}}}{{m - n}}} \right)\)

Note: If P is the mid-point of line segment AB, then \(P\left( {x,\;y} \right) = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\)

CALCULATION:

Here, we have to find the ratio in which the line x + y = 4 divides the line joining the points A (- 1, 1) and B (5, 7).

Let the line x + y = 4 divides the line joining the points A (- 1, 1) and B (5, 7) in the ratio m : 1

Let the point of division be C.

As we know that, the point  internal division is given by: \(\left( {x,\;y} \right) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)

\(⇒ C = \left( {\frac{{5m - 1}}{{m + 1}},\frac{{7m + 1}}{{m + 1}}} \right)\)

∵ C is the point of division i.e C lies on the line x + y = 4 and the coordinates of the point C will satisfy the equation of the line x + y = 4.

\(⇒ \frac{5m - 1}{m + 1} + \frac{7m + 1}{m + 1} =4\)

⇒ (5m - 1) + (7m + 1) = 4(m + 1)

⇒ m = 1/2

So, the required ratio is: (1/2) : 1 = 1 : 2

Hence, option B is the correct answer.

Concept:

1. Direction angles: If α, β, and γ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles.
2. Direction cosines: The cosines of direction angles are the direction cosines of the line. Hence, cos α, cos β and cos γ are called as the direction cosines

It is denoted by l, m and n. ⇔ l = cos α, m = cos β and n = cos γ

1. The sum of squares of the direction cosines of a line is equal to unity.
2. l² + m² + n² = 1 or cos² α + cos² β + cos² γ = 1

Calculation:

Given:

Direction cosines of a line are (1/k, 2/k, -2/k)

So, l = 1/k, m = 2/k and n = -2/k

We know that sum of squares of the direction cosines of a line is equal to unity

⇒ l² + m² + n² = 1

\(\Rightarrow \;\frac{1}{{{k^2}}} + \;\frac{4}{{{k^2}}} + \;\frac{4}{{{k^2}}} = 1\)

\(\Rightarrow \frac{9}{{{k^2}}} = 1\)

⇒ k² = 9

∴ k = ± 3

Concept:

Triangle Law of Vector Addition: When two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

\(⇒ {\rm{\vec R}} = {\rm{\vec A}} + {\rm{\vec B}}\)

Reguar hexagon: A regular hexagon is a kind of polygon with 6 equal sides. 

Properties:

• It is having six sides and six angles.
• The lengths of all the sides are equal.
• Measurements of all the angles are equal.
• The total number of diagonals in it is 9.
• It is having diagonals of two different lengths.
• Short diagonal : d₁ = 3 × √a
• Long Diagonal: d₂ = 2 × a

Solution:

ABCDEF is a Regular hexagon such that \(\vec{AD}\parallel \,\vec{BC}\)  and \(\vec{AD}= 2\vec{BC}\)

By triangle law of vector addition,

\(\vec{AC}=\vec{AB}+\vec{BC}\)

⇒ \(\vec{AC}=\vec{a}+\vec{b}\)

⇒ \(\vec{AC}+\vec{CD}=\vec{AD}\)

⇒ \((\vec{a}+\vec{b})+\vec{CD}=2\vec{b}\)   

⇒ \(\vec{CD}=2\vec{b}-(\vec{a}+\vec{b})\)

⇒ \(\vec{CD}=\vec{b}-\vec{a}\)

⇒ \(|\vec{CD}|= |\vec{FA}|=|\vec{b}-\vec{a}|\)

But the CD and FA are in the opposite direction

⇒ \(- \vec{CD}= \vec{FA}\)

⇒ \(-[\vec{b}-\vec{a}]= \vec{FA}\)

⇒ \(\vec{a}-\vec{b}= \vec{FA}\)

∴ \( \vec{FA}=\vec{a}-\vec{b}\)

Concept:

•  If \(\vec a = \;{a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) is a vector then the magnitude of the vector is given by \(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2} \;\)
• If \(\vec a = \;{a_1}\hat i + {a_2}\hat j + {a_3}\hat k\) is a vector then the unit vector in the direction of \(\vec a\) is given by: \(\hat a = \frac{{\vec a}}{{\left| {\vec a} \right|}}\)
• If A (x, y, z) and B (p, q, r) are two points in a space then the vector \(\overrightarrow {AB} = \left( {p - x} \right)\;\hat i + \left( {q - y} \right)\;\hat j + \left( {r - z} \right)\;\hat k\)

Calculation:

Given: A (1, 2, 3) and B (4, 5, 6) are any two points.

As we know that, If A (x, y, z) and B (p, q, r) are two points in a space then the vector \(\overrightarrow {AB} = \left( {p - x} \right)\;\hat i + \left( {q - y} \right)\;\hat j + \left( {r - z} \right)\;\hat k\)

\(\Rightarrow \overrightarrow {AB} = 3\;\hat i + 3\;\hat j + 3\;\hat k\)

As we know that, if \(\vec a\) is a vector then the magnitude of the vector \(\vec a\) is given by: \(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2} \;\)

Concept:

The angle between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 where a1, b1, c1, a2, b2 and c2 are the direction ratios of the normal to the plane is given by: \(\cos θ = \frac{{|{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}|}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right) ⋅ \left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

Calculation:

Given: The equation of planes is: 2x - 3y + 4z = 1 and - x + y = 4

The given equations can be re-written as: 2x - 3y + 4z - 1 = 0 and - x + y - 4 = 0

By comparing the above equations with a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 , we get

⇒ a1 = 2, b1 = - 3, c1 = 4, a2 = - 1, b2 = 1 and c2 = 0

⇒ a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = - 2 - 3 + 0 = - 5

\(⇒ \sqrt {a_1^2 + b_1^2 + c_1^2} = \sqrt {29} \;and\;\sqrt {a_2^2 + b_2^2 + c_2^2} = \sqrt {2} \)

\(⇒ \cos θ = \; \frac{5}{{\sqrt {58} }}\)

⇒ \(\theta = {\cos ^{ - 1}}\left( { \frac{5}{{\sqrt {58} }}} \right)\)

Concept:

• The shortest distance between the lines \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\)  and \(\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}\) is given by,

\(d =| (\vec A_2 - \vec A_1).\hat {(\vec B_1×\vec B_2)}|\)

\(d =|\frac{ (\vec A_2 - \vec A_1). {(\vec B_1×\vec B_2)}}{|\vec B_1×\vec B_2|}|\)

Where \(\hat {(\vec B_1×\vec B_2)}\) is the unit vector perpendicular to \(\vec B_1 \ and \ \vec B_2 \)

• Here.  \(\vec B_1 \ and \ \vec B_2 \) are representing the direction of the first and second lines \(\vec A_1 \ and \ \vec A_2 \) are representing the points on the first and second lines
• If the two lines are intersecting then,

d = 0

Calculation:

⇒ x1 = t, y1 = 4, z1 = 5  and a1 = 2, b1 = 3, c1 = 4.

⇒ x2 = 0, y2 = 3, z2 = 4  and a2 = 3, b2 = 4, c2 = 5.

\(\vec B_1= (2,3,4)\)

\(\vec B_2 = (3,4,5)\), \(\vec A_1 = (t,4,4)\) and \(\vec A_2 = (0,3,4)\)

\(\vec A_1- \vec A_2 = (t, 1, 1)\)

For intersecting lines,

\(d =|\frac{ (\vec A_2 - \vec A_1). {(\vec B_1×\vec B_2)}}{|\vec B_1×\vec B_2|}| = 0\)

\(⇒ | (\vec A_1 - \vec A_2). {(\vec B_1×\vec B_2)}| = 0\;\;\;\;\; \ldots \left( 1 \right)\)

\(now, \ {(\vec B_1×\vec B_2)} = \begin{vmatrix} \hat i &\hat j &\hat k \\ 2& 3 & 4\\ 3& 4 & 5 \end{vmatrix}\)

\(⇒ {(\vec B_1×\vec B_2)} = (-1, 2, -1)\;\;\;\;\; \ldots \left( 2 \right)\)

By 1 and 2,

⇒ (t, 1, 1) . (-1, 2, -1) = 0

⇒ t  = 1

Hence, option 1 is correct.

Concept:

Let the angle between \(\vec a\) and \(\vec b\)is \(\rm \theta\)

\(\rm \vec a.\vec b = 2ab cos\;\theta\)

Calculations:

consider, the angle between \(\vec a\) and \(\vec b\)is \(\rm \theta\)

Given, \(\vec a + \vec b + \vec c = \vec 0 \)

⇒\(\vec a + \vec b = - \vec c \)

⇒\(\rm |\vec a + \vec b| = |- \vec c |\)

Squaring on both side, we get

⇒\(\rm |\vec a + \vec b|^2 = |- \vec c |^2\)

⇒\(\rm |\vec a|^2 +2\;\vec a.\vec b+ |\vec b|^2 = |- \vec c |^2\)

⇒\(\rm |\vec a|^2 +|\vec b|^2+2\;ab\cos\;\theta = |- \vec c |^2\)

⇒\(\rm (3)|^2 +(5)^2+2\;(3)(5)\cos\;\theta = (7)^2\)

⇒\(\rm 30\cos\;\theta = 15\)

⇒\(\rm \cos\;\theta = \dfrac 12\)

⇒ \(\rm \theta\) = π / 3

Hence, If \(\vec a + \vec b + \vec c = \vec 0,\;|\vec a| = 3,\;|\vec b| = 5\) and \(|\vec c| = 7\), then the angle between \(\vec a\) and \(\vec b\)is π / 3