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Vector Algebra question

2017 · 8 Apr · Shift 1 · Q40
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  5. /2017 · 8 Apr · Shift 1 · Q40

Vector Algebra question

2017 · 8 Apr · Shift 1 · Q40

JEE MainMathematicsVector AlgebraMCQ+4 / −1
The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8i^−6j^8\widehat i - 6\widehat j8i−6j​ and 3i^+4j^−12k^,3\widehat i + 4\widehat j - 12\widehat k,3i+4j​−12k, is :
  1. A
    26
  2. B
    65
  3. C
    20
  4. D
    52
View written solutionFree

Correct answer: B

  1. Let the diagonals of the parallelogram be d⃗1=8i^−6j^+0k^=(8,−6,0),\vec d_1 = 8\hat i - 6\hat j + 0\hat k = (8,-6,0),d1​=8i^−6j^​+0k^=(8,−6,0), d⃗2=3i^+4j^−12k^=(3,4,−12).\vec d_2 = 3\hat i + 4\hat j - 12\hat k = (3,4,-12).d2​=3i^+4j^​−12k^=(3,4,−12).

  2. For a parallelogram with side vectors a⃗\vec aa and b⃗\vec bb, its diagonals are: d⃗1=a⃗+b⃗,d⃗2=a⃗−b⃗.\vec d_1 = \vec a + \vec b, \qquad \vec d_2 = \vec a - \vec b.d1​=a+b,d2​=a−b. The area of the parallelogram is ∣a⃗×b⃗∣.|\vec a \times \vec b|.∣a×b∣.

  3. Now, d⃗1×d⃗2=(a⃗+b⃗)×(a⃗−b⃗).\vec d_1 \times \vec d_2 = (\vec a+\vec b) \times (\vec a-\vec b).d1​×d2​=(a+b)×(a−b). Expanding, =a⃗×a⃗−a⃗×b⃗+b⃗×a⃗−b⃗×b⃗= \vec a\times\vec a - \vec a\times\vec b + \vec b\times\vec a - \vec b\times\vec b=a×a−a×b+b×a−b×b =0−a⃗×b⃗−a⃗×b⃗−0= 0 - \vec a\times\vec b - \vec a\times\vec b - 0=0−a×b−a×b−0 =−2(a⃗×b⃗).= -2(\vec a\times\vec b).=−2(a×b). Therefore, ∣a⃗×b⃗∣=12∣d⃗1×d⃗2∣.|\vec a\times\vec b| = \frac{1}{2}|\vec d_1 \times \vec d_2|.∣a×b∣=21​∣d1​×d2​∣.

  4. Compute d⃗1×d⃗2\vec d_1 \times \vec d_2d1​×d2​:

    \begin{vmatrix} \hat i & \hat j & \hat k \\ 8 & -6 & 0 \\ 3 & 4 & -12 \end{vmatrix}.$$ Expanding, $$= \hat i\big((-6)(-12)-0\cdot 4\big) - \hat j\big(8(-12)-0\cdot 3\big) + \hat k\big(8\cdot 4-(-6)\cdot 3\big).$$ $$= 72\hat i + 96\hat j + 50\hat k.$$

  5. Its magnitude is ∣d⃗1×d⃗2∣=722+962+502|\vec d_1 \times \vec d_2| = \sqrt{72^2+96^2+50^2}∣d1​×d2​∣=722+962+502​ =5184+9216+2500= \sqrt{5184+9216+2500}=5184+9216+2500​ =16900=130.= \sqrt{16900} = 130.=16900​=130.

  6. Hence area of parallelogram =12×130=65.= \frac{1}{2}\times 130 = 65.=21​×130=65.

  7. Therefore the correct option is B: 65.\boxed{\text{B: }65}.B: 65​.

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