Given A=9 i-5 j+2 k, B=11 i+4 j+ k, C=-7 i+14 j-3 k, which statements are true?

GATE 2023 AE – Question 43

Engineering Mathematics · Linear Algebra: Vector algebra, matrix algebra, systems of linear equations, rank, eigenvalues and eigenvectors · 2 marks · Multiple select

Given $\vec A=9\hat i-5\hat j+2\hat k$, $\vec B=11\hat i+4\hat j+\hat k$, $\vec C=-7\hat i+14\hat j-3\hat k$, which statements are true?

A, B and C are coplanar.
Their scalar triple product is zero.
A and B are perpendicular.
C is parallel to $A\times B$.

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Show answer and explanation

Correct answer: (A) A, B and C are coplanar.; (B) Their scalar triple product is zero.

Explanation

**Step 1: cross product** $\vec A\times\vec B$ with $\vec A=(9,-5,2)$ and $\vec B=(11,4,1)$:
$$\vec A\times\vec B=\big((-5)(1)-(2)(4),\ (2)(11)-(9)(1),\ (9)(4)-(-5)(11)\big)=(-13,\ 13,\ 91).$$

**Step 2: scalar triple product** with $\vec C=(-7,14,-3)$:
$$(\vec A\times\vec B)\cdot\vec C=(-13)(-7)+(13)(14)+(91)(-3)=91+182-273=0 .$$

**A. Coplanar.** Zero triple product means the three vectors are coplanar. ✓
**B. Triple product is zero.** ✓
**C. A and B are perpendicular.** $\vec A\cdot\vec B=99-20+2=81\neq0$. ✗
**D. $\vec C$ is parallel to $\vec A\times\vec B$.** $(-7,14,-3)$ is not a multiple of $(-13,13,91)$. (Since $\vec C$ lies in the plane of A and B, it is actually perpendicular to $\vec A\times\vec B$.) ✗

Answer **A and B**.