Various Forms Of The Equation Of A Line

Let the points be and mid point of
and mid point of
We know, mid point of both the diagonal lie on the same point .
and
...(i)
and ... (ii)
Slope of slope of

... (iii)
Solving Eqs. (ii) and (iii), we get

Now, slope of slope of




The required points are and .

Let . Then, So, Thus, Therefore,

We are given the limit expression: To solve this, we first expand around using the Taylor series: Substitute this expansion into the expression : Now, substitute this into , and expand the cosine term similarly: Using the Taylor expansion for cosine again: Simplifying: Now, substitute this into the original limit expression: This simplifies to: Thus, the value of the limit is , making the correct answer Option B.

We are given the equation of the plane: The equation of the line is given in symmetric form: Step 1: Direction ratios of the line We can write the direction ratios of the line from the symmetric form: Step 2: Normal to the plane The normal to the plane is given by the coefficients of and in the plane equation: Step 3: Condition for parallelism For the plane and the line to be parallel, the direction ratios of the line must be perpendicular to the normal vector of the plane. The condition for perpendicularity is that the dot product of the direction ratios of the line and the normal vector of the plane must be zero: Taking the dot product: Simplifying: Step 4: Find We are asked to find . From the equation , divide through by 2: Now, multiply both sides by 3: Thus, the correct answer is Option B.

Consider a system of equations that has a non-trivial solution, implying that the determinant of the system's matrix must be zero. Given the matrix we apply column operations and to simplify the determinant: Expanding along , we calculate: Dividing by and , we simplify: This results in the relation: confirming that for a non-trivial solution to exist, specific relations between the parameters must hold. Thus, the correct answer is Option B.

Step 1: Identify the equation and its roots. The quadratic equation given is . By Vieta's formulas, the sum of the roots and (where and are roots) is given by: Step 2: Expressing in terms of . Since , we have: Step 3: Verifying the correct option for . Upon checking each option with the value of , option C is confirmed where: which simplifies correctly under the assumption that satisfies the original equation, and any transformations follow algebraic rules that apply to the equation's roots.

We are given the expression: Let . The conjugate of is . Thus, we can rewrite the expression as: This expression simplifies as: Step 1: Take the power of the expression Now, we need to find the smallest positive integer such that: is purely imaginary. Using the properties of complex exponentiation: For the expression to be purely imaginary, the real part must be zero, i.e., This occurs when , or . Thus, the smallest positive integer such that the expression is purely imaginary is . Thus, the correct answer is Option A.

This leads to the calculation:
So, the correct answer is Option D.

Starting with the equation given: This simplifies as: Which further reduces to: Expanding this, we find: Solving for , we equate to the general solution of trigonometric equations: Ensuring does not exceed 18, we calculate the sum: