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Q.(a) Equation of the plane with intercepts 2, 3, 4 on the x, y and z axis respectively is

(i) 2x + 3y + 4z = 1
(ii) 2x + 3y + 4z = 12
(iii) 6x + 4y + 3z = 1
(iv) 6x + 4y + 3z = 12 (Score : 1)
(b) Find the Cartesian equation of the plane passing through the points A(2, 5, -3), B(-2, -3, 5) and C(5, 3, -3). (Scores : 3)
Kerala DhseKerala DHSE Plus Two Board 2016Subjective· 4mImportance★★★★★
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The intercept form xa+yb+zc=1\frac xa+\frac yb+\frac zc=1 gives the plane equation for given intercepts; the plane through three non-collinear points is found using the normal vector AB⃗×AC⃗\vec{AB}\times\vec{AC}.

(a) Intercept form with a=2, b=3, c=4a=2,\ b=3,\ c=4: x2+y3+z4=1.\dfrac x2+\dfrac y3+\dfrac z4=1. Multiply by 1212 (LCM of 2,3,42,3,4): 6x+4y+3z=12.6x+4y+3z=12. Answer: (iv).

(b) A(2,5,−3), B(−2,−3,5), C(5,3,−3).A(2,5,-3),\ B(-2,-3,5),\ C(5,3,-3).

AB⃗=B−A=(−4,−8,8),AC⃗=C−A=(3,−2,0).\vec{AB}=B-A=(-4,-8,8),\quad \vec{AC}=C-A=(3,-2,0).

Normal n⃗=AB⃗×AC⃗=∣ı^ȷ^k^−4−883−20∣\vec n=\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\-4&-8&8\\3&-2&0\end{vmatrix}

=ı^[(−8)(0)−(8)(−2)]−ȷ^[(−4)(0)−(8)(3)]+k^[(−4)(−2)−(−8)(3)]=\hat\imath[(-8)(0)-(8)(-2)]-\hat\jmath[(-4)(0)-(8)(3)]+\hat k[(-4)(-2)-(-8)(3)]

=ı^(0+16)−ȷ^(0−24)+k^(8+24)=16ı^+24ȷ^+32k^.=\hat\imath(0+16)-\hat\jmath(0-24)+\hat k(8+24)=16\hat\imath+24\hat\jmath+32\hat k.

Divide by 88: n⃗=(2,3,4)\vec n=(2,3,4). …

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