(a) If the given vectors are orthogonal, their dot product is zero.
1
𝛼
Hence we have 3 (17𝛼 − 17) + 3 (17𝛼 2 ) + 2 (−17𝛼 +
17
)
6
=0
This gives (17𝛼 − 17) + 17𝛼 3 − 102𝛼 + 17 = 0
On further simplification, we get 17𝛼 3 − 85𝛼 = 0. But 𝛼 is given to be non zero, we get 𝛼 2 − 5 = 0
Hence the value of 𝛼 is given by ±√5
(b) Any vector perpendicular to two given vectors a and b is parallel to the vector a x b. So, we can
take t(a x b) as some vector perpendicular to both a and b. Here, the given vectors are <0, 42, 42> and
<𝜋, 0, −𝜋 >. The direction ratios of these vectors are <0, 1, 1,> and <1, 0, -1>. The vector perpendicular
𝑖 𝑗 𝑘
to these two vectors is parallel to |0 1 1 | = −𝑖 + 𝑗 − 𝑘.
1 0 −1
Now let the required vector is 𝑡(−𝑖 + 𝑗 − 𝑘). But given that the required vector has length equal to 1.
Hence we have √𝑡 2 + 𝑡 2 + 𝑡 2 = 1. This implies 𝑡 = ±
Hence the required vector is ±
1
(−𝑖
√3
1
.
√3
+ 𝑗 − 𝑘)
(c) The given equation does not imply x=y. The reason: 𝑥 × 𝑣 = 𝑦 × 𝑣 ⇒ (𝑥 − 𝑦) × 𝑣 = 0. This
implies that the vector (𝑥 − 𝑦) is either zero or parallel to the vector 𝑣. So, the second possibility makes
the conclusion 𝑥 = 𝑦 wrong. It is in fact, one of the two possibilities only.