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3D Geometry question

2022 · 26 Jun · Shift 2 · Q35
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  5. /2022 · 26 Jun · Shift 2 · Q35

3D Geometry question

2022 · 26 Jun · Shift 2 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1
Let a→=i^+j^+2k^\overrightarrow a = \widehat i + \widehat j + 2\widehat ka=i+j​+2k, b→=2i^−3j^+k^\overrightarrow b = 2\widehat i - 3\widehat j + \widehat kb=2i−3j​+k and c→=i^−j^+k^\overrightarrow c = \widehat i - \widehat j + \widehat kc=i−j​+k be three given vectors. Let v→\overrightarrow vv be a vector in the plane of a→\overrightarrow aa and b→\overrightarrow bb whose projection on c→\overrightarrow cc is 23{2 \over {\sqrt 3 }}3​2​. If v→ . j^=7\overrightarrow v \,.\,\widehat j = 7v.j​=7, then v→ . (i^+k^)\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)v.(i+k) is equal to :
  1. A
    6
  2. B
    7
  3. C
    8
  4. D
    9
View written solutionFree

Correct answer: D

  1. Write v⃗\vec vv as a linear combination of a⃗\vec aa and b⃗\vec bb since it lies in the plane of a⃗\vec aa and b⃗\vec bb:

v⃗=xa⃗+yb⃗\vec v = x\vec a + y\vec bv=xa+yb

Given a⃗=(1,1,2),b⃗=(2,−3,1),c⃗=(1,−1,1)\vec a=(1,1,2),\quad \vec b=(2,-3,1),\quad \vec c=(1,-1,1)a=(1,1,2),b=(2,−3,1),c=(1,−1,1)

So, v⃗=x(1,1,2)+y(2,−3,1)=(x+2y, x−3y, 2x+y)\vec v = x(1,1,2)+y(2,-3,1)=(x+2y,\ x-3y,\ 2x+y)v=x(1,1,2)+y(2,−3,1)=(x+2y, x−3y, 2x+y)

  1. Use the condition v⃗⋅j^=7\vec v\cdot \hat j=7v⋅j^​=7.

Since j^=(0,1,0)\hat j=(0,1,0)j^​=(0,1,0), this means the yyy-component of v⃗\vec vv is 777:

x−3y=7(1)x-3y=7 \qquad (1)x−3y=7(1)

  1. Use the projection condition.

The scalar projection of v⃗\vec vv on c⃗\vec cc is

v⃗⋅c⃗∣c⃗∣=23\frac{\vec v\cdot \vec c}{|\vec c|}=\frac{2}{\sqrt 3}∣c∣v⋅c​=3​2​

Now, ∣c⃗∣=12+(−1)2+12=3|\vec c|=\sqrt{1^2+(-1)^2+1^2}=\sqrt 3∣c∣=12+(−1)2+12​=3​

Hence,

\implies \vec v\cdot \vec c=2$$ Compute: $$\vec v\cdot \vec c=(x+2y,\ x-3y,\ 2x+y)\cdot(1,-1,1)$$ $$=(x+2y)-(x-3y)+(2x+y)=2x+6y$$ So, $$2x+6y=2 \implies x+3y=1 \qquad (2)$$ 4. **Solve** equations (1) and (2): $$(1)\; x-3y=7$$ $$(2)\; x+3y=1$$ Adding, $$2x=8 \implies x=4$$ Then from (2), $$4+3y=1 \implies 3y=-3 \implies y=-1$$ 5. **Find** $\vec v$: $$\vec v=4\vec a-\vec b$$ $$=4(1,1,2)-(2,-3,1)=(4,4,8)-(2,-3,1)=(2,7,7)$$ 6. **Compute** $$\vec v\cdot (\hat i+\hat k)=(2,7,7)\cdot(1,0,1)=2+7=9$$ Therefore, the required value is $$\boxed{9}$$ So the correct option is **D**.
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