Question 12 - Jee advanced Math 2022 P1 Questions with Solutions

Let S be the reflection of point Q with respect to the plane given by

\(\vec{r} = -(t + p)\hat{i} + t\hat{j} + (1 + p)\hat{k}\)

where t, p are real parameters and \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along the three positive coordinate axes. If the position vectors of Q and S are \(10\hat{i}+ 15\hat{j} + 20\hat{k}\) and \(\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}\) respectively, then which of the following is/are TRUE?

A) \(3(\alpha + \beta) = -101\)

B) \(3(\beta + \gamma) = -71\)

C) \(3(\gamma + \alpha) = -86\)

D) \(3(\alpha + \beta + \gamma) = -121\)

Sol : 

S is the reflection of Q.

\(\implies\) Both are same distance(perpendicular) from the given plane.

Drop perpendiculars from S and Q onto the plane; let A be the point on the plane where they meet.



Let \((x_{1}, y_{1}, z_{1})\) be the coordinates of A.

\(\implies\) \(\overrightarrow{OA} = x_{1}\hat{i} + y_{1}\hat{j} + z_{1}\hat{k}\) is position vector of A.

Also, let \(\overrightarrow{OQ} = 10\hat{i}+ 15\hat{j} + 20\hat{k}\) and \(\overrightarrow{OS} = \alpha \hat{i}+ \beta \hat{j} + \gamma \hat{k}\).

Define vector \(\overrightarrow{AQ}\) from A to Q, and vector \(\overrightarrow{AS}\) from A to S. Both these are vectors perpendicular to the plane.

By triangle law of vector addition, 

\(\overrightarrow{AS} = \overrightarrow{OS} - \overrightarrow{OA}\)

         \( = \alpha \hat{i}+ \beta \hat{j} + \gamma \hat{k} - (x_{1}\hat{i} + y_{1}\hat{j} + z_{1}\hat{k})\)

         \(= (\alpha - x_{1})\hat{i} + (\beta - y_{1})\hat{j} + (\gamma - z_{1})\hat{k}\)

But vectors \(\overrightarrow{AS}\) and \(\overrightarrow{AQ}\) are equal and opposite.

\(\implies \overrightarrow{AS} = - \overrightarrow{AQ}\)

\(\implies - \overrightarrow{AQ} = (\alpha - x_{1})\hat{i} + (\beta - y_{1})\hat{j} + (\gamma - z_{1})\hat{k}\)

\(\implies \overrightarrow{AQ} = (x_{1} - \alpha)\hat{i} + (y_{1} - \beta)\hat{j} + (z_{1} - \gamma)\hat{k}\)

\(\implies \overrightarrow{OQ} - \overrightarrow{OA} = (x_{1} - \alpha)\hat{i} + (y_{1} - \beta)\hat{j} + (z_{1} - \gamma)\hat{k}\) ….{by triangle law of vector addition}

\(\implies (10 - x_{1})\hat{i} + (15 - y_{1})\hat{j} + (20 - z_{1})\hat{k} = (x_{1} - \alpha)\hat{i} + (y_{1} - \beta)\hat{j} + (z_{1} - \gamma)\hat{k}\)

Two vectors are equal \(\implies\) their components are equal.

\(\implies \alpha = 2x_{1} - 10\),

         \(\beta = 2y_{1} - 15\) and

        \(\gamma = 2z_{1} - 20\).

Now all we need to do is find \(x_{1}, y_{1}, z_{1}\) and substitute them back in these equations.

The given parametric equation of the plane is
\(\vec{r} = -(t + p)\hat{i} + t\hat{j} + (1 + p)\hat{k}\)

Rewriting the equation as:
\(\vec{r} = 0\hat{i} + 0\hat{j} + 1\hat{k} + t(-\hat{i} + \hat{j} + 0\hat{k}) + p(-\hat{i} + 0\hat{j} + \hat{k})\)
where 
\(0\hat{i} + 0\hat{j} + \hat{k}\) is the position vector of the point (0, 0, 1) in the plane and
\(-\hat{i} + \hat{j} + 0\hat{k}\) and \(-\hat{i} + 0\hat{j} + \hat{k}\) are vectors in the plane.

Also, \(\overrightarrow{AQ} = (10 - x_{1})\hat{i} + (15 - y_{1})\hat{j} + (20 - z_{1})\hat{k}\) is a vector perpendicular to the plane, hence also perpendicular to the vectors \(-\hat{i} + \hat{j} + 0\hat{k}\) and \(-\hat{i} + 0\hat{j} + \hat{k}\).

\(\implies (10 - x_{1})(-1) + (15 - y_{1})(1) + (20 - z_{1})(0) = 0\) and
      \((10 - x_{1})(-1) + (15 - y_{1})(0) + (20 - z_{1})(1) = 0\).

\(\implies x_{1} - y_{1} + 5 = 0\) and
      \(x_{1} - z_{1} + 10 = 0\).

Let letter B denote the point (0, 0, 1) on the plane. Define \(\overrightarrow{BA}\). This is also a vector in the plane.



\(\overrightarrow{BA} = (x_{1} - 0)\hat{i} + (y_{1} - 0)\hat{j} + (z_{1} - 1)\hat{k}\)

From the equation, \(-\hat{i} + \hat{j} + 0\hat{k}\) and \(-\hat{i} + 0\hat{j} + \hat{k}\) are vectors in the plane, hence their cross product \((-\hat{i} + \hat{j} + 0\hat{k}) \times (-\hat{i} + 0\hat{j} + \hat{k})\) will be a vector perpendicular to the plane. 

\(\implies \overrightarrow{BA} \cdot [(-\hat{i} + \hat{j} + 0\hat{k}) \times (-\hat{i} + 0\hat{j} + \hat{k})] = 0\)



\(\implies (x_{1} - 0)(1) + (y_{1} - 0)(1) + (z_{1} - 1)(1) = 0\)
\(\implies x_{1} + y_{1} + z_{1} = 1\)

And from earlier we had these two equations
\(x_{1} - y_{1} + 5 = 0\) and
 \(x_{1} - z_{1} + 10 = 0\)

Adding these two will give us
\(2x_{1} - (y_{1} + z_{1}) + 15 = 0\)
\(\implies 2x_{1} - (1 - x_{1}) + 15 = 0\)
\(\implies x_{1} = -\frac{14}{3}\)
\(\implies y_{1} = -\frac{14}{3} + 5 = \frac{1}{3}\) and 
             \(z_{1} = -\frac{14}{3} + 10 = \frac{16}{3}\)

Substituting the values of \(x_{1}, y_{1}, z_{1}\) in the equations for \(\alpha, \beta, \gamma\):

\(\alpha = 2(-\frac{14}{3}) - 10 = -\frac{58}{3}\),

\(\beta = 2(\frac{1}{3}) - 15 = -\frac{43}{3}\) and

\(\gamma = 2(\frac{16}{3}) - 20 = -\frac{28}{3}\)

Let’s check the options now.

A) \(3(\alpha + \beta) = -101\)

     \(\implies 3(\frac{- 58 - 43}{3}) = -101\)

    \(\implies -101 = -101\) which is true

B) \(3(\beta + \gamma) = -71\)

   \(\implies 3(\frac{- 43 - 28}{3}) = -71\)

    \(\implies -71 = -71\) which is true

C) \(3(\gamma + \alpha) = -86\)

    \(\implies 3(\frac{- 28 - 58}{3}) = -86\)

    \(\implies -86 = -86\) which is true

D) \(3(\alpha + \beta + \gamma) = -121\)

     \(\implies 3(\frac{- 58 - 43 - 28}{3}) = -121\)

    \(\implies - 129 = -121\) which is not true

\(\implies\) Options A, B and C are True.

Comments

Post a Comment

Popular posts from this blog

Question 7 - Jee advanced Math 2022 P2 Questions with Solutions

Image
 Consider the hyperbola \(\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1\) with foci at \(S\) and \(S_{1}\),  where \(S\) lies on positive x-axis. Let \(P\) be a point on the hyperbola, in the first quadrant. Let \(\angle{SPS_{1}} = \alpha\), with \(\alpha < \frac{\pi}{2}\). The straight line passing through the point \(S\) and having the same slope as that of the tangent at \(P\) to the hyperbola, intersects the straight line \(S_{1}P\) at \(P_{1}\). Let \(\delta\) be the distance of \(P\) from the straight line \(SP_{1}\), and \(\beta = S_{1}P\). Then the greatest integer less than or equal to \(\frac{\beta \delta}{9}\sin{\frac{\alpha}{2}}\) is ______. Sol :  There are a couple of properties of hyperbola that will help us here. (1) The tangent line at a point bisects the angle between the lines connecting the two foci with the point of tangency.  So, in the figure above, tangent at \(P\) bisects the angle \(\alpha\) between the lines joining the foci(\(S\) an...
Read more

Question 6 - Jee advanced Math 2022 P2 Questions with Solutions

 Let \(\beta\) be a real number. Consider the matrix \(\Biggl(\matrix{\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2}\Biggl)\). If \(A^{7} - (\beta - 1)A^{6} - \beta A^{5}\) is a singular matrix, then the value of \(9\beta\) is _____. Sol:  \(A^{n}\) is a matrix A multiplied to itself n times. Consider \(A^{7}\)            \( = A^{6} \cdot A\)            \( = (A^{5} \cdot A) \cdot A\)            \( = A^{5} \cdot (A \cdot A)\)….. Associative law of matrix multiplication            \( = A^{5} \cdot A^{2}\) \(A^{7} - (\beta - 1)A^{6} - \beta A^{5}\)  \( = A^{5} \cdot A^{2} - (\beta - 1)[A^{5} \cdot A] - \beta A^{5}\) For any constant \(p\) and square matrices \(A\) and \(B\),  \(p (A \cdot B) = (pA) \cdot B = A \cdot (pB)\)  \(\implies = A^{5} \cdot A^{2} - A^{5} \cdot [(\beta - 1) A] - \beta A^{5}\) Also, for any square mat...
Read more

Question 3 - Jee advanced Math 2022 P1 Questions with Solutions

In a study about a pandemic, data of 900 persons was collected. It was found that 190 persons had symptom of fever, 220 persons had symptom of cough, 220 persons had symptom of breathing problem, 330 persons had symptom of fever or cough or both, 350 persons had symptom of cough or breathing problem or both, 340 persons had symptom of fever or breathing problem or both, 30 persons had all three symptoms(fever, cough and breathing problem). If a person is chosen randomly from these 900 persons, then the probability that the person has at-most one symptom is _____. Sol :  Let C be a letter for representing ‘Cough’; F for ‘Fever’; and B for ‘breathing problem’.  i) 190 = n(F) = n(F only) + n(F and C) + n(F and B) + n(all three) ii) 220 = n(C) = n(C only) + n(F and C) + n(C and B) + n(all three) iii) 220 = n(B) = n(B only) + n(F and B) + n(C and B) + n(all three) n(F or C or both) = n(F) + n(C) - n(F and C) - n(all three) (We are subtracting n(F and C) because it is already in...
Read more