Skip to content
Question 150 of 162

Q.(a) Find the parametric form of Vector equation and Cartesian equations of the plane containing the line r⃗=(i^−j^+3k^)+t(2i^−j^+4k^)\vec{r}=(\hat{i}-\hat{j}+3\hat{k})+t(2\hat{i}-\hat{j}+4\hat{k}) and perpendicular to the plane r⃗⋅(i^+2j^+k^)=8\vec{r}\cdot(\hat{i}+2\hat{j}+\hat{k})=8. OR

(b) Solve the equation 6x4−5x3−38x2−5x+6=06x^4-5x^3-38x^2-5x+6=0 if it is known that 13\dfrac13 is a solution.
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2023Subjective· 5mImportance★★★★★
93% · 150/162 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

(a) Builds the required plane from a point and two direction vectors — the line's direction and the given plane's normal (since the planes are perpendicular) — then takes their cross product for the normal; (b) uses the given root together with the palindromic symmetry of the quartic's coefficients to factor and solve completely. Both alternatives answered below.

(a) Plane containing the line, perpendicular to the given plane

1. Data from the line. The line r⃗=(i^−j^+3k^)+t(2i^−j^+4k^)\vec r=(\hat i-\hat j+3\hat k)+t(2\hat i-\hat j+4\hat k) passes through point A(1,−1,3)A(1,-1,3) with direction d⃗1=(2,−1,4)\vec d_1=(2,-1,4) — both lie in the required plane.

2. Perpendicularity condition. The required plane is perpendicular to r⃗⋅(i^+2j^+k^)=8\vec r\cdot(\hat i+2\hat j+\hat k)=8, whose normal is n⃗2=(1,2,1)\vec n_2=(1,2,1). When two planes are perpendicular, the normal of one plane is a direction lying within the other plane. So d⃗2=(1,2,1)\vec d_2=(1,2,1) is a second direction vector of our plane.

3. Normal of the required plane. n⃗=d⃗1×d⃗2=∣i^j^k^2−14121∣\vec n=\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&4\\1&2&1\end{vmatrix}

i^[(−1)(1)−(4)(2)]−j^[(2)(1)−(4)(1)]+k^[(2)(2)−(−1)(1)]=i^(−1−8)−j^(2−4)+k^(4+1)=−9i^+2j^+5k^\hat i[(-1)(1)-(4)(2)]-\hat j[(2)(1)-(4)(1)]+\hat k[(2)(2)-(-1)(1)]=\hat i(-1-8)-\hat j(2-4)+\hat k(4+1)=-9\hat i+2\hat j+5\hat k

So n⃗=(−9,2,5)\vec n=(-9,2,5).

4. Vector (parametric) equation.

r⃗=(i^−j^+3k^)+t(2i^−j^+4k^)+s(i^+2j^+k^),t,s∈R\vec r=(\hat i-\hat j+3\hat k)+t(2\hat i-\hat j+4\hat k)+s(\hat i+2\hat j+\hat k),\quad t,s\in\mathbb R

5. Cartesian equation. Using point (1,−1,3)(1,-1,3) and normal (−9,2,5)(-9,2,5):

−9(x−1)+2(y+1)+5(z−3)=0 ⇒ −9x+9+2y+2+5z−15=0 ⇒ −9x+2y+5z−4=0-9(x-1)+2(y+1)+5(z-3)=0\ \Rightarrow\ -9x+9+2y+2+5z-15=0\ \Rightarrow\ -9x+2y+5z-4=0

i.e. 9x−2y−5z+4=09x-2y-5z+4=0. Check: point (1,−1,3)(1,-1,3): 9(1)−2(−1)−5(3)+4=9+2−15+4=09(1)-2(-1)-5(3)+4=9+2-15+4=0 ✓.

(b) Solve 6x4−5x3−38x2−5x+6=06x^4-5x^3-38x^2-5x+6=0, given x=13x=\dfrac13 is a root

1. Coefficients are palindromic (6,−5,−38,−5,66,-5,-38,-5,6 read the same forwards and backwards), which is the signature of a reciprocal equation — if rr is a root, so is 1r\dfrac1r. Since 13\dfrac13 is a root, x=3x=3 is also a root.

2. Quadratic factor from these two roots. (x−13)(x−3)(x-\tfrac13)(x-3), scaled to integer coefficients as (3x−1)(x−3)=3x2−10x+3(3x-1)(x-3)=3x^2-10x+3.

…

Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous questionQ51 — (a) Using vector method, prove that cos(α-β)=cosαcosβ+sinαsinβ **OR** …→Next questionQ53 — If a vector vecα lies in the plane of vecβ and vecγ, then : (a) [vecα,…