Vector Algebra
Cross Product and Scalar Triple Product
Grade 12
Question:
<p>Let \(a\), \(b\), \(c\) be three vectors such that \([a b c] = 2\). If \(r = l(b \times c) + m(c \times a) + n(a \times b)\) is perpendicular to \(a + b + c\), then the value of \((l + m + n)\) is</p>
<p>(a) 2</p>
<p>(b) 1</p>
<p>(c) 0</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The perpendicularity condition combined with scalar triple product properties forces the coefficients to sum to zero when the scalar triple product is non-zero.
Step 1: Since \(r\) is perpendicular to \(a + b + c\), we have: \[r \cdot (a + b + c) = 0\] Step 2: Substituting \(r = l(b \times c) + m(c \times a) + n(a \times b)\): \[[l(b \times c) + m(c \times a) + n(a \times b)] \cdot (a + b + c) = 0\] Step 3: Expanding using properties of scalar triple product: \[l[(b \times c) \cdot a] + l[(b \times c) \cdot b] + l[(b \times c) \cdot c] + m[(c \times a) \cdot a] + m[(c \times a) \cdot b] + m[(c \times a) \cdot c] + n[(a \times b) \cdot a] + n[(a \times b) \cdot b] + n[(a \times b) \cdot c] = 0\] Step 4: Most terms vanish (product of cross product with parallel vectors = 0). We get: \[l[a b c] + m[a b c] + n[a b c] = 0\] Step 5: Since \([a b c] = 2 \neq 0\): \[l + m + n = 0\]
Correct Answer: C