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Vector Algebra question

2023 · Shift 2 · Q21
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Vector Algebra question

2023 · Shift 2 · Q21

JEE AdvancedMathematicsVector AlgebraMCQ+3 / −1
Let the position vectors of the points P,Q,RP, Q, RP,Q,R and SSS be a⃗=i^+2j^−5k^,b⃗=3i^+6j^+3k^\vec{a}=\hat{i}+2 \hat{j}-5 \hat{k}, \vec{b}=3 \hat{i}+6 \hat{j}+3 \hat{k}a=i^+2j^​−5k^,b=3i^+6j^​+3k^, c⃗=175i^+165j^+7k^\vec{c}=\frac{17}{5} \hat{i}+\frac{16}{5} \hat{j}+7 \hat{k}c=517​i^+516​j^​+7k^ and d⃗=2i^+j^+k^\vec{d}=2 \hat{i}+\hat{j}+\hat{k}d=2i^+j^​+k^, respectively. Then which of the following statements is true?
  1. A
    The points P,Q,RP, Q, RP,Q,R and SSS are NOT coplanar
  2. B
    b⃗+2d⃗3\frac{\vec{b}+2 \vec{d}}{3}3b+2d​ is the position vector of a point which divides PRP RPR internally in the ratio 5:45: 45:4
  3. C
    b⃗+2d⃗3\frac{\vec{b}+2 \vec{d}}{3}3b+2d​ is the position vector of a point which divides PRP RPR externally in the ratio 5:45: 45:4
  4. D
    The square of the magnitude of the vector b⃗×d⃗\vec{b} \times \vec{d}b×d is 95
View written solutionFree

Correct answer: B

  1. Write the given position vectors
a⃗=(1,2,−5),b⃗=(3,6,3),c⃗=(175,165,7),d⃗=(2,1,1)\vec a=(1,2,-5),\quad \vec b=(3,6,3),\quad \vec c=\left(\frac{17}{5},\frac{16}{5},7\right),\quad \vec d=(2,1,1)a=(1,2,−5),b=(3,6,3),c=(517​,516​,7),d=(2,1,1)

We must check which option is true.

  1. Check whether P,Q,R,SP,Q,R,SP,Q,R,S are coplanar

The four points are coplanar iff the scalar triple product of

PQ→=b⃗−a⃗,PR→=c⃗−a⃗,PS→=d⃗−a⃗\overrightarrow{PQ}=\vec b-\vec a, \quad \overrightarrow{PR}=\vec c-\vec a, \quad \overrightarrow{PS}=\vec d-\vec aPQ​=b−a,PR=c−a,PS=d−a

is zero.

Compute:

PQ→=(3−1,6−2,3−(−5))=(2,4,8)\overrightarrow{PQ}=(3-1,6-2,3-(-5))=(2,4,8)PQ​=(3−1,6−2,3−(−5))=(2,4,8) PR→=(175−1,165−2,7−(−5))=(125,65,12)\overrightarrow{PR}=\left(\frac{17}{5}-1,\frac{16}{5}-2,7-(-5)\right)=\left(\frac{12}{5},\frac{6}{5},12\right)PR=(517​−1,516​−2,7−(−5))=(512​,56​,12) PS→=(2−1,1−2,1−(−5))=(1,−1,6)\overrightarrow{PS}=(2-1,1-2,1-(-5))=(1,-1,6)PS=(2−1,1−2,1−(−5))=(1,−1,6)

Now,

PR→×PS→=∣i^j^k^12565121−16∣\overrightarrow{PR}\times \overrightarrow{PS} = \begin{vmatrix} \hat i & \hat j & \hat k\\ \frac{12}{5} & \frac{6}{5} & 12\\ 1 & -1 & 6 \end{vmatrix}PR×PS=​i^512​1​j^​56​−1​k^126​​ =i^(65⋅6−12(−1))−j^(125⋅6−12⋅1)+k^(125(−1)−65⋅1)=\hat i\left(\frac{6}{5}\cdot 6-12(-1)\right) -\hat j\left(\frac{12}{5}\cdot 6-12\cdot 1\right) +\hat k\left(\frac{12}{5}(-1)-\frac{6}{5}\cdot 1\right)=i^(56​⋅6−12(−1))−j^​(512​⋅6−12⋅1)+k^(512​(−1)−56​⋅1) =i^(365+12)−j^(725−12)+k^(−125−65)=\hat i\left(\frac{36}{5}+12\right)-\hat j\left(\frac{72}{5}-12\right)+\hat k\left(-\frac{12}{5}-\frac{6}{5}\right)=i^(536​+12)−j^​(572​−12)+k^(−512​−56​) =(965,−125,−185)=\left(\frac{96}{5},-\frac{12}{5},-\frac{18}{5}\right)=(596​,−512​,−518​)

Then

PQ→⋅(PR→×PS→)=(2,4,8)⋅(965,−125,−185)\overrightarrow{PQ}\cdot (\overrightarrow{PR}\times \overrightarrow{PS}) =(2,4,8)\cdot \left(\frac{96}{5},-\frac{12}{5},-\frac{18}{5}\right)PQ​⋅(PR×PS)=(2,4,8)⋅(596​,−512​,−518​) =1925−485−1445=0=\frac{192}{5}-\frac{48}{5}-\frac{144}{5}=0=5192​−548​−5144​=0

So the points are coplanar. Hence:

  • Option A is false.
  1. Compute b⃗+2d⃗3\dfrac{\vec b+2\vec d}{3}3b+2d​
b⃗+2d⃗3=(3,6,3)+2(2,1,1)3=(3,6,3)+(4,2,2)3=(7,8,5)3=(73,83,53)\frac{\vec b+2\vec d}{3} =\frac{(3,6,3)+2(2,1,1)}{3} =\frac{(3,6,3)+(4,2,2)}{3} =\frac{(7,8,5)}{3} =\left(\frac73,\frac83,\frac53\right)3b+2d​=3(3,6,3)+2(2,1,1)​=3(3,6,3)+(4,2,2)​=3(7,8,5)​=(37​,38​,35​)

Let this point be XXX.

  1. Check whether XXX divides PRPRPR internally in the ratio 5:45:45:4

For a point dividing PRPRPR internally in the ratio 5:45:45:4,

x⃗=5c⃗+4a⃗5+4\vec x=\frac{5\vec c+4\vec a}{5+4}x=5+45c+4a​

or equivalently depending on convention,

x⃗=5a⃗+4c⃗9\vec x=\frac{5\vec a+4\vec c}{9}x=95a+4c​

We test the option by direct computation.

Using the standard section formula for a point dividing PRPRPR in ratio 5:45:45:4 means

x⃗=5r⃗+4p⃗9=5c⃗+4a⃗9\vec x=\frac{5\vec r+4\vec p}{9}=\frac{5\vec c+4\vec a}{9}x=95r+4p​​=95c+4a​

Now,

5c⃗=(17,16,35),4a⃗=(4,8,−20)5\vec c=\left(17,16,35\right), \qquad 4\vec a=(4,8,-20)5c=(17,16,35),4a=(4,8,−20)

So,

5c⃗+4a⃗=(21,24,15)5\vec c+4\vec a=(21,24,15)5c+4a=(21,24,15)

Thus

5c⃗+4a⃗9=(219,249,159)=(73,83,53)\frac{5\vec c+4\vec a}{9}=\left(\frac{21}{9},\frac{24}{9},\frac{15}{9}\right) =\left(\frac73,\frac83,\frac53\right)95c+4a​=(921​,924​,915​)=(37​,38​,35​)

which is exactly

b⃗+2d⃗3\frac{\vec b+2\vec d}{3}3b+2d​

Hence this point divides PRPRPR internally in the ratio 5:45:45:4.

So:

  • Option B is true.
  • Option C is false.
  1. Check option D

We need ∣b⃗×d⃗∣2|\vec b\times \vec d|^2∣b×d∣2.

b⃗×d⃗=∣i^j^k^363211∣\vec b\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 6 & 3\\ 2 & 1 & 1 \end{vmatrix}b×d=​i^32​j^​61​k^31​​ =i^(6⋅1−3⋅1)−j^(3⋅1−3⋅2)+k^(3⋅1−6⋅2)=\hat i(6\cdot 1-3\cdot 1)-\hat j(3\cdot 1-3\cdot 2)+\hat k(3\cdot 1-6\cdot 2)=i^(6⋅1−3⋅1)−j^​(3⋅1−3⋅2)+k^(3⋅1−6⋅2) =3i^+3j^−9k^=3\hat i+3\hat j-9\hat k=3i^+3j^​−9k^

Therefore,

∣b⃗×d⃗∣2=32+32+(−9)2=9+9+81=99|\vec b\times \vec d|^2=3^2+3^2+(-9)^2=9+9+81=99∣b×d∣2=32+32+(−9)2=9+9+81=99

So option D is false.

  1. Final conclusion

The only true statement is:

B\boxed{\text{B}}B​

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